Then and then the state said you paid me 4,000, but I'm going to give you 2,000 back Well, that means that you didn't really pay 4,000 because they refunded 2,000 So what should you do? Should I go back to the prior year tax return and put only 2,000 here? Which is actually the proper amount I paid after the refund was refunded back to me That would be tedious to do or we could do a cashed based system and try to fix it going forward And that's why you might have to include in income the refund that you get in the following period So that's going to be how that that that kind of plays out now Remember that that does get a lot more confusing than you would think at first glance meaning if you got a tax refund Of $2,000 because you paid 4,000 in the prior year Then the question is well if you didn't itemize last year Then you didn't get any benefit and you wouldn't have to include it in income if you did itemize It's still questionable as to whether you've got a benefit From the taxes or how much of a benefit you got because notice that that my standard deduction right here Is is 12,950 So what if my itemized deductions just barely cleared that what if my itemized deductions came out to be like 13,000 or something Then I just barely cleared the 12,940 so even though I got a $4,000 state tax deduction And I was itemizing I really only got maybe $50 of it right because I because the 12,950 I would have gotten any case so it gets messy in terms So that's why the tax software is quite useful to determine If you got a benefit in that case and that's why I would suggest if you are dealing with a new client That had a schedule a in the prior year that it's worthwhile to Data input the prior year tax return in the prior year software even though it might cost more Roll it over so that the software can help you out with some of those rollover situations Like the refund situation now a similar thing could happen for The other taxes like the like if you have the sales tax Then you might have to include it in other income like down here If you've got a refund a situation for the other taxes that you paid But that's a little bit more unusual of a situation now You also might think well that would lead people possibly To try to like you could try to say if I had a lot of income this year And my tax bracket is going to be higher you could you know people may say well I'm going to try to withhold a lot This year so that so that my it lowers my income this year And I'll be at higher tax brackets this year because my income was higher And then when I get the refund next year, I'll include it in income next year And and because I'm at lower tax brackets next year It'll be a net benefit people can try to play a little manipulative games like that with the cutoff And you got and and the reason That's the problem with a cashed based kind of system. So you got to you got to be You got to be careful whenever that idea of that idea comes into play Well, I'm going to I'm going to prepay all my expenses This year or something sometimes the tax code will limit you in terms of how much you can kind of prepay But it's they have a little bit of a difficulty doing that with the state taxes Because the state tax has a progressive tax system So it's quite likely that people have no idea how much to pay for the taxes Because it's a pay-as-you-go system and a progressive tax system So so there's that all right, and then we also could have included in here You got to be careful of the cutoff kind of situations because If if for example I made estimated tax payments or say I got a refund From my state taxes last year that I rolled over roll over Or into the current year because I'm a Schedule C business Then then that is something that will typically be included in the taxes as well So let's take a look at that So if I had my estimated tax payments, so now I'm not having payments Necessarily withheld from the W2 although I still have those But what if I need to make estimated tax payments as well Which are often done with like a Schedule C business if there was an over payment applied from 2021 And you apply it then to the taxes Let's say it was another 1,000 Then that is going to be included in here as well. So now my state taxes went up to 5,000 So so that's something you've got to you've got to kind of be aware of As well if you're not a W2 employee or if you have the kind of cutoff situation between When you paid the tax and what year it should be applied to Also, if I made a payment in the current year for taxes That were for 2021 for prior year taxes, but I paid them in 2022 That may be included in the Schedule A because again, they're tied to generally a cash based system when you paid them Not an accrual based system when the tax year That's being applied to So for example, if you paid with 2021 state tax return A thousand dollars, let's say Then that was applied to tax year 2021 But you paid it in 2022. So it increased my amount here by 6,000 by a thousand again to 6,000 So it's on it's basically generally on kind of a cash based system And again, if you're using tax software, it will usually help you to pick that stuff up But then you want to kind of double check it if you're looking at this number And if it's you want to say, how is the system calculating? That if it's coming from the Sales the the the Taxes your income taxes you paid for the state Then it would be easy if it just came from the W2, but you also want to think about that cutoff Kind of stuff. Did I have a refund last year? Did I owe money last year if I owe money in 2021 and I paid it In 2021 then I I paid for 2021 in 2022 when I filed the 2021 tax return by april 15th 2022 Then that might be included in here as well and you want to get that that kind of cutoff problem Situated so then you can have the personal property taxes This is often with like the dmv Like vehicle tax and whatnot and you've got to make sure that you take into consideration the the deductible part Of the value of it So I'll just say 160 and pull that on over So that's going to be important. It's usually going to be a relatively small of course In comparison to the property tax and the state taxes, but could still be relevant worth picking up, of course


Some of you are so afraid of losing the Holy Spirit for every mistake that you make. Galatians 5.16 shows us that the Holy Spirit is our power unto holiness. I promise you, God would not remove from you your only chance for being holy as a punishment for you not being holy. What am I saying to you that you can go on sinning by no means? I'm saying the Holy Spirit faithfully abides with you to help you get it right.


Okay, let's go ahead and get started then. Thanks everybody for making it out. This week we've got Miko Spona from the University of Florida who will be talking to us about non-rational generating. Take us away. Hi, thank you for the invitation. I love to go to the University of South Carolina. This is my fourth time that the last two times were conferences in 2015 and 2009, but my first time and also the last time when I presented something at this seminar was in mid-November of 2000, and which was a very similar time to now because right now we don't know who the next president will be and back then we didn't know who the next president will be even though it was in the middle of November in case you remember the Bush versus Gore time. Okay, so you can see my shared screen, right? So I'm going to talk about non-rational generating functions and then there will be a particularly interesting to me class of examples. So let's see, why can't I go to the next? Why can't I go to the next? I guess I can go manually like this. Okay, so in this class, sorry, in this talk when I say generating function, I always mean the ordinary generating function. Okay, so just this thing. This talk, there will be no exponential or other generating functions. So here is the main idea of the talk that sometimes we have an enumeration problem and we can't solve it. It's, if we try and we can't solve it, we are trying to get an exact formula and we cannot find it. And then we don't know that we don't find it because we are just not good enough or because the problem is actually difficult, okay? And sometimes one thing that helps that can actually give you some kind of reassurance that the problem is actually difficult is when you can say something about the generating function of this enumeration problem, even though you can't find it in its closed form because if you could then you would have solved the problem, but you can say something about it that shows that it's not an ice generating function. That usually indicates that it's not an easy enumeration problem because if it was easy then there would be a nice answer to it and the generating function of that nice answer would be nice too. Okay, so I will give a brief introduction to the parts of generating functionality that we need today. It's maybe 10, 15 minutes. I apologize to the experts like Gabor, you can go to sleep, but just wake up after 10 minutes. All right, so yes, because I don't know that the whatever function here doesn't work on my computer so you will see some things twice. So for as far as niceness of generating functions is concerned, there will be three classes of generating functions in terms of how nice they are, which are positioned like this. Rational functions are the nicest, right? The ratio of two polynomials. Algebraic generating functions, those that solve an algebraic equation are not quite as nice but still nice and then there are defined generating functions. I will tell you what that is, which are not as nice as the previous two but still it's something. If your generating function is defined or different should be finite then there is still a chance for some kind of an ice formula. Okay, so the first class is rational and generating functions. The SI set is the ratio of two polynomials. And this is equivalent to the fact that the coefficients of your power series, so these are usually the numbers that count objects of size n, satisfy a recurrence relation with constant coefficient. These numbers are constants, okay? So for instance, fn is equal to seven times f9 plus three times f8. fn is equal to seven times fn minus one plus two times fn minus two, something like this. Okay, these are the rational functions. So that's the nicest. It's often easy to find a solution for that reason because you look for a recurrence relation with constant coefficients. Sorry about this. I can't, it seems that I cannot go from one page to another, so the pause function of slides. Ah, okay, now I've made it something, oops. No, it still doesn't work. So I just have to manually go from one page to another. I don't know why that is. Okay, so we were at rational. Okay, so then the next class, a little bit wider class of power series is algebraic generating functions. Algebraic power series. So we say a power series is algebraic if it solves a polynomial equation like this. So if a power series is not algebraic, then it's unlikely that you will have a super simple formula for the coefficients, but there can still be a formula. Just it won't be, it will contain more than just the four basic operations and roots. Also the other problem with algebraic generating functions that there is not an easy, sufficient and necessary condition for the coefficients. So remember, above I said that if a power series is rational, then its coefficients satisfy a recurrence relation with constant coefficients. For algebraic power series, I cannot say such an easy equivalence. Okay, everything that's rational is also algebraic, namely of degree one, because then it satisfies a polynomial equation with degree one, never multiply your rational function with its denominator and you get the numerator. And then there are those defined or differentially finite power series. Okay, so the way to define this is the following. Take all the derivatives of your power series and that certainly spans a vector space over the field of rational functions.  So for instance, E to the Z is D finite because all these derivatives are E to the Z. So that vector space will have dimension one, cosine Z because it's every other derivative is cosine and every other derivative is sine. So it will have the vector space will have dimension two or ln one plus Z. None of these are algebraic as it can be proved, but they are all D finite. So on the other hand, it can be proved. It's not completely trivial, but it's not difficult either that algebraic power series are D finite. It's in the Stanley book, Enumerative Combinatorics, volume two, chapter six. So D finite power series are bigger. That class of D finite power series properly contains that of algebraic power series. Now, remember I said that for algebraic power series, it wasn't easy to say what is true for the coefficients, but for D finite sequences, it is. So a sequence is called polynomial recursive. So it's my sequence. It's not the power series, it's a sequence of numbers. It's called polynomial recursive or p recursive. If it satisfies a linear recurrence relation this time with polynomial coefficients, okay? So for rational functions, the coefficients in the recursion had to be constants. Here they have to be polynomials, okay? So there is at least this aspect in which D finite power series are nicer than algebraic ones. So you could say at this point, sometimes people ask if I can give an example of a problem whose solution, of a simple problem whose solution is a sequence which is not polynomial recursive or whose generating function is not D finite. And my answer is the bell numbers. The number of all partitions of the n elements set, right? It's generating function is e to the x minus one. And the bell numbers do satisfy a recurrence relation, but it's not with a constant number of terms. This k is, you can't fix that. So that keeps growing. The bn can be obtained as n choose i times b i and sum that for i goes from zero to n. And that's not a lot here. Here this k has to be fixed. Okay, so all right. These are the three classes of power series I wanted to talk about. That's in order to put the rest of my talk into context. Okay, so now back to my main topic of non-rational generating functions. So let's say that you have an enumeration problem that you can't solve. And you wonder that maybe this is difficult and maybe I could prove that it's difficult by showing that it's, at least the generating function is not very nice. I mean, it would be great as if you could prove that the generating function is not even D finite, but that is very difficult because there are very few methods to prove that something is not D finite. Maybe I will talk about that later. So how do you prove that something is not rational, that the generating function is not rational? How do you prove it if you don't know the generating function itself? You don't know the generating function itself, but how can you maybe know enough about it to prove that it's not rational? So what we will use is what Flajole and Sej recall the main law of combinatorial asymptotics. Okay, so this is important. If you have a combinatorial generating function, meaning that all the coefficients are non-negative real numbers, then the exponential growth rate of the coefficients is equal to one over R where R is the radius of convergence of F around zero. So you kind of learn this in calculus, but this is what we will need today. In combinatorial generating functions, there are nice properties sometimes that are not true in calculus in general. Okay, so the notion that you might not know is super criticality. This goes as far as Sino first introduced in the Flajole-Sejvik book, no, maybe it was in some papers before, but there the theory is developed. So that's the super criticality that's a property of a relation between two generating functions, okay? So let F and G be two generating functions with non-negative real coefficients that are analytic at zero. Let's say that G zero is zero. And let's say that F and G are related to each other like this. I will mention soon that this is extremely frequent in combinatorics. This happens all the time. The two generating functions are related to each other like this. Then this relation is called super critical if when you put in the radius of convergence of G into G itself, what you get is bigger than one, okay? So let's see what this means. It has an analytic meaning and a combinatorial meaning. So what is the analytic meaning? So I put in the radius of convergence here and if it's bigger than one, that means that sometime earlier it was equal to one, right? Because G is a combinatorial generating function. So if you put in a smaller number then you get a smaller number. So it's monotone in Z. So if G at RG is larger than one then there's a smaller number of alpha where G of alpha is equal to one. Hence this denominator is zero. Hence this function F has a singularity. So F has a singularity before G does. So in other words, the radius of convergence of F is smaller than the radius of convergence of G. And therefore the exponential growth rate of the coefficients of F is larger than that of G, right? Because I said that the exponential growth rate was the reciprocal of the radius of convergence. So if F and G are connected by this very frequent relation, then the coefficients of F have a larger growth, strictly larger exponential growth than those of G. Is this clear? Everybody's on mute. Could I please get a verbal yes? Yes. That's good. Thank you. Okay, so now what does this mean combinatorially? Combinatorially F, so I said that this happens all the time. F equals one over one minus G means that one is equal to, that F is equal to one plus all the powers of G. So if F counts structures that can have several components and each component is counted by G and the order of the components doesn't, in the order of components is, it matters but the components have to be, they cannot intermingle. So they have to be like interval. You cut up an interval into smaller intervals. Okay, so if F and G are combinatorial structures so that F is put together by components which are G, then if the relation is super critical, then F grows faster than G. And the exponential growth rate of F is larger than the exponential growth rate of G. Those are some examples and then in two minutes I will start my main talk. Okay, so for instance, if F is the generating function for the number of ways to tie that one times N hat with red and blue times of size one times one and three kinds of times of size one times two, then GZ has this polynomial generating function implying that its convergence reduces infinity. So of course G at infinity is bigger than one. And therefore the relation is super critical and indeed F has this generating function which in that the coefficients can be, you can compute that the growth rate of the coefficients, it's something like three, but it's certainly not zero. In G the growth rate of the coefficients is zero because there are no coefficients larger than the second one. But here the coefficient, this has growth rate about three to the N. Another example is let F be the number of ways to split a semester into N days of non-empty segments and then choosing one day for a class social in each segment, then GZ describes what you can do in one segment. If your segment has M days, then you can make M choices. So this is your generating function for G. Okay, clearly this has growth, exponential growth rate one, but F has this generating function which has exponential growth rate 2.618. Let's declare something. This has to do with the Fibonacci numbers. Okay, but now let me give you a famous example there which is not super critical and we are getting close to a real subject of the talk. Well, okay, that's gonna come too but so sorry, there's difference like coming first. If G is rational, then this relation is super critical. Okay, so you can see that I'm getting with this. In order to prove that something is not rational, it would be enough to prove that this relation is not super critical. And that we will do later today, right? Because if G is that, let's prove this theorem first. So there are two cases, either G is a polynomial or it's not. If G is a polynomial, then it's convergent everywhere. The convergence rate uses infinity. So G at infinity is of course bigger than one. So yes, that's the condition for super criticality. So the relation is super critical. Otherwise, if G is a rational function that is not the polynomial, it has at least one singularity and all those singularities are poles. What do we know about poles, right? And from complex analysis that around the pole, and this is important, this will be the punchline later that around the pole the function goes to infinity, which is not true for other type of singularities. So not necessary through any way. So, right, so G at RG is infinity, which is again bigger than one, completing the proof that this relation has to be super critical in if F and G are rational. In fact, if one of them is rational, then the other is necessary for rational, okay? So therefore, as I said, if you want to prove that some generating function about which you don't know much is not rational, it's enough to prove that this relation is not super critical. And the way to do that, one other way to do that is to show that F and G have the same convergence radius or the coefficients have the same exponential order, right? Because we argued before that if the relation is super critical, then F grows faster than G. So if they have the same growth rate that put their coefficients, then the relation cannot be super critical, so the functions cannot be rational. Okay, that could be my punchline. So a famous example of a non-supercritical relation is the Catalan numbers, where F is this famous generating function, which can be obtained this way from G, where G happens to be Z times FZ. And because, in this case, G is just Z times FZ, wherever F has a singularity, G has a singularity. So they both have dominant singularities at one-fourth. So the relation is not super critical because F and G have the same exponential order, the coefficients have the same exponential order, F and G have the same dominant singularity, whereas if the relation was super critical, then F would have a smaller dominant singularity. All right. Okay, so that was the intro talk. So if you were sleeping, this is a very good time to wake up. Okay, now I start applying what I said. Permanent mutation patterns will be my main example because I work on them a lot. They are widely studied. Their generating functions are mostly unknown. There are just a very, very few sporadic cases when you actually know what the generating function is. And there is a long-standing, difficult conjecture about their generating functions, which I will not solve, but I will go a long way. I mean, long way, I will make a step towards it. Okay, so this, the next two minutes are maybe the most important for the rest of the talk, but I'm just trying to get to a figure here, which is worth 10,000 pictures. So yeah, if you came with the idea that you will pay attention for five minutes, then please make it these five minutes because otherwise the rest will be unenjoyable. So we say that the long permutation like P contains a smaller permutation in this case, Q. If you can choose entries in the big one, which don't have to be in consecutive positions, but they relate to each other the same way as the entries of the small one. So five, eight, six, that's the same pattern as one, three, two, because the first entry is the smallest, the second entry is the largest, and the third entry is the middle one in size. So in this case, I can say that this long permutation P decreasing case. So for short patterns that is if Q is of length three, then it is known, it is known, it's not trivial, but it's like an advanced undergraduate exercise or beginning graduate exercise that if Q is of length three, any one of the six patterns of length three, then the generating function, then the number of permutations of length N avoiding Q is the Catalan number, which has this generating function, which is algebraic. After that, so once Q is longer than three, it's no longer true that the number of permutations avoiding Q doesn't depend on Q itself. So it's no longer true that Q can be any pattern of length four, for instance, and A and Q will always be the same. It's very far from being true. Otherwise, there wouldn't be this field of pattern avoiding permutations. So here is one example. This is my result from 1996. The story of this is that I already handed in my thesis and then I found this result. So I withdrew the thesis so I could put this in. I was so happy about it. So if Q is this pattern of length four, then this is the generating function, which is again algebraic, not rational but algebraic. By the 24 patterns of length four, there are only three that are inequivalent and this is one of them. This is the biggest equivalence class and the only one for which the exact form, well, kind of, the only one for which the generating function is known in this sense that I can describe it like that. Okay, I said that the other exception when something is known about the generating functions of pattern avoiding permutations is when the pattern is monotone, then it is known that the generating function is differentially finite. However, if K is larger than two and it's even then it's not algebraic. So in particular for length four, one, two, three, four, the generating function is not algebraic and that's surprising, right? Because I could tell the generating function of this one. So the monotone pattern is not the nicest because this one has an algebraic generating function and the monotone one doesn't. Okay, how do we prove that those generating functions are not algebraic? Because we have a good estimate of their size. And there is a theorem that says that if a power series is algebraic and the growth rate of the coefficients is alpha to the n over n to the K, then that K has to be an integer and a half. It cannot be an integer and that would happen in the case when K is even and more than two. So those functions don't have the right growth rate and that's why they are not algebraic. And that's about the only way that we can tell that something is not algebraic for as far as pattern avoiding permutations. Otherwise, of course, there are other results. Okay, so what do we know about the size of pattern avoiding permutations? There's a celebrated theorem. It was the Stanleyville conjecture was opened for 23 years and then Adam Marcus and Gabor Tardos solved it without actually knowing that they were solving it. They proved that this number, the number of permutations of length n that the point of given pattern Q is less than a constant to the n. They solved the theorem, a fairly hard enough question from matrix, extrema matrix theory, and that happened to imply this result. So this result was very difficult to prove because if you think about it, this is a very, very ambitious statement, right? There are n factorial permutations of length n and only a constant to the n of them avoids Q. So very, very little. So most methods of most probabilistic methods are not fine enough to prove that something is that small. Nobody knows what the best value of that constant is. There were conjectures and then they were disproved. The conjecture somehow said the value of the best constant should be a polynomial function of the length of the pattern. But that was disproved in 2013 by Jacob Fox who showed that there exist patterns where the constant is an exponential function of the pattern line. So now we don't even know what the conjecture. Yes, there are explicit formulas for these two in terms of numbers, but that for the monotone one doesn't mean a nice generating function. Okay, all right. Yeah, I said that, so yeah. And so, yeah, let me tell you this because this will be another punchline. So remember I said before that for patterns of length four, there are really only three non-equivalent cases, this one, this one and one, three, two, four. For these two, we have an exact formula for the number of, for these numbers. However, for one, three, two, four, four it is wickedly difficult. So we don't even know the exponential growth rate better than a factor of two and 2.6 or something. So we know it's bigger than 10.8 and smaller than 13.5. So this is just a really, really difficult question. Doron Seiberger says that maybe even God doesn't know what is the value of A1, A1,000, one, three, two, four. Also, people ran also sorts of gigantic programs which showed that if the generating function for this pattern is a d finite, then it has to be d finite with some horribly high degree, like more than 500. Remember d finite means the finite dimensional vector space. So that vector space would have a huge dimension. Okay. So there was the Seiberger Newton conjecture in 1996 and I will talk about the single pattern case. So let, again, let's recall that this is the number of permutations that avoid Q. And the conjecture was that for all Q, that this is a single pattern version. This sequence is polynomial recursive or equivalently the generating function of this difference should be finite. Note that if instead of avoiding one pattern, you want to avoid multiple patterns, then the conjecture was disproved by Igor Puck and Scott Garabrant maybe four years ago or so. Their counter examples are huge, like 500 patterns of length 80 or something, 500 patterns of length 80 or something like that. But for the single pattern case, when you want to avoid one pattern, this is still open. So the single pattern, again, the conjecture is that, or the conjecture was that the generating function of for the number of permutations avoiding a single pattern will be d finite, differentially finite, equivalently, the numbers will form a polynomial recursive sequence. And so what I will do in this talk is that no, I can't prove, I can't decide a question whether they are d finite or not, but I can prove that for the overwhelming majority of patterns, the generating function will be not rational. Okay, so for all patterns except like one over and square of them. And also it will be funny to see what's the first permutation for which I cannot decide a question. Okay, so we will prove in two different ways that this is not rational. In fact, I will prove a little bit more that they are not rational. I will also prove that the dominant singularity is not a pole, right? It could be technically that they are not rational, but the non-pol singularity arrives somewhere later, but no, I will prove that the dominant singularity is not a pole, hence they cannot be rational. All right, so here is a simple definition. We say that the permutation is skew in the composable if you cannot cut it into two parts so that everything before the cut is larger than everything after the cut. For instance, three, one, four, two is skewing the composable, but three, four, six, five, one, two is not because you can cut it say after five so that everything before the cut is bigger than everything after the cut, okay? If P is not skewing the composable then you can cut it in a unique way to non-empty skewing the composable string, strings, so that in each of them, the entries are bigger than in the next one. These I will call the skew blocks, okay? Here will be some examples. This has four skew blocks, right? In each block, the entries are bigger than in the next block, but each block itself is skewing the composable. It can't be cut any further, okay? So here is my first theorem and I will later I will make it a little bit stronger but this is the main push. So let Q be skewed, so it is missing, let Q be skewing in the composable pattern that does not end in its largest entry. Unfortunately, I have to say that. You will see that that's important and otherwise it's not even true. Well, I will talk about that later. So if Q is like that, it's skewing the composable and it does end in its largest entry, then its generating function is not rational. By the way, almost four permutations of a given length are skewing the composable and don't end in their largest entry. So these are numerically not big restrictions. So if this happens, then AQ is not rational. The proof is remarkably simple for a theorem of this strength and then it's even more annoying that for those remaining small cases, the remaining few cases we don't know and we didn't know now for almost two years. Lots of people wanted to do it after my paper came out and then it's still unknown. So it's probably difficult. So clearly, because Q is skewing the composable, P avoids Q if each of the blocks avoid it, right? If each of the skew blocks of P avoids Q, right? Because that's the point that if Q is skewing the composable, you cannot start a copy here and continue it here because that would make it skew the composable, okay? So this means that you can build up the generating function AQ like this, right? Q avoiding permutation can be put together from skew in the composable Q avoiding permutations so that you place them like this. Hence, this generating function identity holds. You are seeing where I'm getting with this, right? That's exactly what will lead to the question of supercriticality. So remember that we said that if A is rational, then this relation should be supercritical so these two sequences could not have the same exponential growth rate. But I will show that they do and so that we'll prove that they are not supercritical and so they cannot be rational. And this is quite easy to do actually. Let P be your flanked N and let it avoid Q. Now I fix the new N3 N plus one at the end, okay? So let's go back to my picture. Here is P, each of these blocks avoid Q so the whole P avoids Q. Now I put a new big entry N plus one here. So that new permutation of flanked N plus one still avoids Q because that last entry N plus one is the maximal entry. It could only play the role of the maximal entry in any perforated copies of Q. But remember in my theorem, I said that Q does not end in its largest entry, okay? So in other words, that new permutation of flanked N plus one that I created still avoids Q, hence this in a chain of inequalities forms, right? The first one is trivial. Again, at this N1 that means skewing the composable. It has one skew block, okay? And so on the other hand, because in the, on the other hand, the new permutation that I created by putting an N plus one at the end is of course skewing the composable because it ends in N plus one, right? N plus one must always be in the first skew block. So this is true in other words, the exponential growth rate of this sequence on the right is not smaller than the exponential growth rate of the sequence in the middle. Therefore these two sequences have the same exponential order. And so this one on the left also has the same exponential order because it's the same sequence. The sequences on the left end and the right end are the same sequences just shifted by one. So this contradiction completes the proof that they cannot be, they cannot have the same exponential order, right? So now I know that this generating function is not rational as long as Q is skewing the composable and it does not end in its largest entry. Now, if skew is not skewing the composable, so if that's the problem then you turn it backwards, right? You take the reverse of it. Then the reverse of it is automatically skewing the composable and clearly the number of permutations avoiding P, the number of permutations avoiding Q. So avoiding Q and avoiding Q's reverse are the same. So that takes care of that problem that in the composability problem. Now, how about the problem that it has to end with the largest entry, right? So there you can do some trickery. So if you have the permutation that ends in the largest entry then you can take its reverse complement. So you reverse the permutation and then you subtract each entry from N plus one, right? And then you play with this a little bit and then you see that the only time you cannot avoid, you can't defeat the conditions by the story taking reverses and taking complements is the following case. When you have a pattern that starts with one, sorry. So your theorem will be true for all patterns which are skewing the composable such that at least one of the following conditions fold. It doesn't start with the entry one. It doesn't end with the entry K. Or if both of those are bad, like if you take the monotone pattern, one, two, three, four, five, six, then what if it's will effect equivalent? And I will tell you what that is to skewing in the composable pattern that satisfies at least one of the first conditions. So we have two patterns are called will effect equivalent if they are avoided by the same number of permutations of length N for every N. And there are theorems which say for instance that one, two, three, so the monotone pattern is avoided by here, so the monotone pattern is will effect equivalent to the one where you flip the first two entries, okay? And then the second one already is fine because it doesn't start with one. So therefore my theorem of non-rationality holds for the monotone pattern because it holds for this pattern two, one, three, two, one, and then monotone increasing to K. Okay, and as I said again, if your pattern is not skewing the composable then you can just reverse it, so that will be fine too. So that you do have a problem when your pattern starts with one and ends with the maximal entry and it's not will effect equivalent to anything else. And unfortunately, I mean, I don't know unfortunately or not, but you should have a problem there because the theorem is false for the trivial problem, the trivial pattern one, two, right? If a permutation avoids the pattern one, two, that means that it's the monotone decreasing permutation. There is one such permutation for every n, so the generating function is this and so it's obviously rational. But is this the only exception? Probably, but we can't prove it. So what is the first time that I cannot decide the question for a pattern? So the first entry has to be one, the last entry has to be the maximum entry K and it cannot be the monotone pattern because for those I can, so that the smallest you can do is one, three, two, four. See, again, that wicked pattern, that one for which we can't say anything about the size. But that's also the smallest one for which we cannot decide rationality even though people strongly believe that the generating function will not even be finite, let alone algebraic, let alone rational but we cannot prove rationality. So that innocent-looking fact that it ends with its largest entry and starts with the smallest entry is we couldn't beat that and not just me, many people try it. Let me show you a slightly different proof because it has its own advantages and it's not very difficult from here. So let's denote by A and two, the number of permutations of flanked N that avoid Q which have two skew blocks. These are those which are skewing the composable so they have one skew block, okay? So if AQ1 is rational, then it's dominant singularity. The singularity that's closest to zero is at pole, okay? So if it's a pole, then around the pole as Z approaches that singularity, the function converges to infinity or diverges to infinity. So if A, which is a monotone increasing function converges to infinity at that pole, then somewhere before that, there's a point where it's bigger than one, right? All the coefficients are non-negative. Sometime before you reach infinity, you will get bigger than one but you will still be finite. You will still be within the convergence radius. There, a contradiction occurs. What do we know about numbers that are bigger than one? What we know is that if you square them, they go up, okay? So their square is bigger than themselves. So take the square of your power series, your generating function for skewing the composable permutations. That gives you the generating function for permutations avoiding Q that have two skew blocks. So that would mean that this number is bigger than this number, right? If you put in the same Z1 into this generating function, you get a bigger number than when you put it into this generating function. And that's a contradiction because these are both power series with non-negative integer coefficients, yes? And I said that An1 is always as large as An2. Again, that can be proved very similarly to what I have done before by sticking a big entry at the end. So that's a contradiction because this long chain of inequalities shows that the left is at least as large as the right. So that there are more, right? Whereas this says the opposite. So it cannot be that you have more powers, you have more objects with one component than two components and have a rational generating functions for that. With generating functions for that. Both of those things cannot happen. If you have a generating function that is rational, then there would have to be more objects with two components than with one, okay? How could we prove the missing kids? So I couldn't, here are some, some atoms that might work or I couldn't make them work. Here is another interesting fact about super critical structures. So again, just think about super critical structures as one kind of object is put together by the other kinds of object as components. It is known, it's in the flagellate book, for instance, that in that case, the expected number of components is linear, okay? So translating this to the language of pattern avoiding permutations, if the generating function would be rational, then there would be a linear, on average, there would be a linear number of skew in the composable components. And that sounds really, really false because in the case is when we know that expectation, it's not only not linear, but it's a constant, okay? It would be enough to prove that the expected number of components is like n to the 0.99. And that would be enough to show that the generating function is not rational, but we can do that. This is what I said before that it proved that the dominant singularity is not a pole, right? Because I only argued about the dominant singularity and I proved it wasn't a pole. Because if it was then, yeah, then I could find a place where the generating function was bigger than one. So in fact, I proved even a little bit more than that. I proved that at that singularity, the value of your generating function has to be less than one. Here is another approach, which is very annoying that I couldn't make it work because it would be a nice example. So note that our proof works for this pattern, one, two, four, five, three, because it doesn't end in the largest entry, but it doesn't work for this one because it starts with the smallest and with the largest. So, numerical evidence suggests that more permutations avoid this one for which the theorem works, the theorem works, then this one for which the theorem doesn't. Unfortunately, I can't prove that, but let's say that I could. Then I could prove, this is not obvious, but I could prove that the two generating functions would have the same convergence radius. And then I would put in that same convergence radius in those two generating functions. And if indeed the coefficients on the left are bigger than the coefficients on the right, then this would have, this would follow. Uh-huh. But the left-hand side is finite because for that, my theorem works, right? And so, so remember I said that when you put in the dominant singularity in a generating function for which the theorem works, you have to get a number less than one. So that most one. So this would be a number which is at most one. So this would be a number which is at most one. So that R could not be a pole for this generating function, AQ2, right? So that that would work nicely. Finally, and this is the last thing I will just mention that in his book, Animative combinatorics volume two, Richard Stanley lists six general families of combinatorial objects, very wide ones that are counted by the same sequences and shows that those sequences are always algebraic with some very basic assumptions. So yeah. Now lattice paths, polynomial dissections, some threes, that sort of thing. And just because there's a very wide class of generating functions, which is algebraic, then you could think that occasionally they will be rational because an algebraic generating function typically contains a big square root and under this or other root. And occasionally, if your class is wide enough, it could happen that those, what happens to be say, under a square root sign will actually be a perfect square. And then your generating function could by accident turn out to be rational. But with this method, I can prove that that's not the case. We are short on time. So I will not go through the details, but by the same kind of argument, there is a lattice path and I turn it into another lattice path which is in the composable because it never touches the baseline. So none of those functions, which are like the go-to-place for algebraic generating functions, none of those are rational. So this is all I wanted to say. I hope you found it interesting. And I also hope at some point life becomes normal and I can visit South Carolina again. It doesn't have to be in a presidential election year. Thank you. If we could all take our speaker in some way, then we'll open it up for questions. Okay, do we have any questions for our speaker? Oh, sorry, I haven't looked at the chat now. Oh, okay. Just three thank yous, okay. So, okay, no questions. So I'm certainly looking forward to your next physical visit to Columbia. Thanks. I would like to ask a question that there are some hundred years old problem on which really nothing happened. This is like the stamp folding problem. So you have a horizontal string of stamps and there are perforations between them. And the question would be that how many ways you can fold this string along the perforations to a single rectangle shape of a stamp. And it has an even nastier version which is the map folding problem that everybody knows whoever the overcar and using a map. Is there anything known on those problems that are hard and they have no nice generating functions? I don't know. Thank you. I don't have to look at it because I think in this method, there is a lot more than it was used for. So I will look into that. So the stamp folding problem. Stamp folding and map folding. Do I have any other questions? Okay. In that case, thanks again. Be close and have a good weekend everybody. Bye. Bye. Bye.


Hi, my name is Catherine. I'm the Music and Fine Arts Librarian at the University of Alabama. This video is an introduction to finding scores and audio recordings of music with the Library's Discovery Tool Scout. One of the advantages of using Scout is that it allows us to simultaneously search the library's catalog electronic resources and databases. Starting at the library's homepage, lib.ua.edu, select the Scout link from the quick menu. This will take you to the basic search. In Scout, we can search by keyword, phrase, title, author, or a combination of several search terms. For example, let's say we want to find a score of Aaron Copeland's Lincoln Portrait for Speaker and Orchestra. We're going to start by searching for the composer's name in the title of the piece. Since we want to search for the title as a phrase, we're going to add quotation marks around Lincoln Portrait. When we run the search, Scout returns a large number of hits, in this case over 2,000. This includes resources like academic journal articles, audio recordings, eBooks, periodical scores, and more. In order to narrow our results by format, we can use the filtering options on the left side of the results page. Scroll down to Source Types and select Show More to view the complete list. From this list, we can select Source Types by checking the corresponding boxes. Selecting music scores will limit our results to only 9 hits, including both print and electronic resources. This item is an electronic version of the score that can be accessed by selecting electronic item available online. Click here. We also have a print score that can be checked out from the library. The library catalog information on this page tells us the score can be found in the music library, located on the first floor of the Gorgas Library. The call number is also listed for locating the item on the shelf, and we can see that the item is available for checkout. We can also select the title to view the full Scout record, which will contain additional information about the resource. To find a recording, we can change our Source Type to Audio. Our search results contain a number of streaming recordings that can be accessed online, like this one from the Naxos Music Library. In Scout, we can use additional information in our search, such as the names of conductors, performers, ensembles, and record labels to narrow our results. We can also add terms like CD, compact disc, or LP to search for a particular format. That's all for now. Thank you for watching. If you have any questions or would like help using UA Libraries, contact us or visit ask.lib.ua.edu to ask a librarian.


Hello, good afternoon. So first of all, thank you for RubyConf talent for giving me this opportunity to talk here. So today, I would like to talk about security issues on your Python, oh, sorry, on your Ruby code. And let me introduce myself. I'm Harley Davidson. It sounds weird. It sounds like the motorbike. And I'm working as security program manager for application security firm. Singapore based, but I'm working for Jakarta region. It's called Funtouch Point Security. So I manage how the client rolling out the application security program end to end from the requirement design, development, and until to the deployment. So coming from Jakarta Indonesia, you can reach out to me at Twitter, at Harley-David, car one. No wonder it's hard for me to use Harley-Davidson in the Twitter account because it's already used. And today, I'm going to tell a story about the finding security issue on development stage using open source static application security testing. Is there anyone familiar about the term? So is there anyone coming from the security background? Yeah, we have one there. OK, so I'm going to talk about the static application security testing in the back. So the goal is the developer can identify security issues on earlier stage without waiting for application through penetration testing. Is there anyone knows or ever heard about penetration testing? Yeah, a lot. So penetration testing is a stage where we literally hack our application to measure to find whether the app contains security issues or not. And this is about the common development process. The common development process starts from the requirement, design development testing in UAT and after that going through the production stage. Ideally, in every stage of a development stage, ideally there must be security touch point involved in every stage of a development stage. Because like I mentioned earlier about the penetration testing, if we take a look on this picture, penetration testing is taking part in the end of the development stage. Means we need to wait for the application being created before we're doing some security assessment. Meanwhile, security issues could be created on the earlier stage, such as on design phase and development phase as well. And what happened if we only put security assessment on the final stage? After we found the security issues, we need to roll back the process coming through from the requirement, doing some design, and develop again. And after we fixing for the first time, there is no security issue there and there is no security measurement. And again, on the final stage, we're doing penetration testing and then we found the security issue and then it's coming back to the first stage. So it's kind of like circle loop of the remediation cycle. There is a research artifact which said that the comparison of cost to fix security issue based on time of detection. So it's the research coming from NISD. So NISD said that if we found the security issues on the acceptance testing, it will be 15 much expensive rather than if we found immediately on the earlier stage. So it's for the management point of view perspective. It will be costly hire if we put security assessment in the final stage of development because we need to spend more budget to redevelop, to redesign, and fix all of the stuff. And I was mentioned about security issues. So in my opinion or in some of literature said that security issues coming from two different kind. The first one, the coming from the design flows. Design flows like there's no RBIAC implementation. Like let's say the regular employee able to invoke the supervisor function in the application, something like that. And the other one is coming from the coding box. So today we're going to talk more deep about the coding box. Not really deep, actually. So SQL injection is one of them. Any familiar about SQL injection? Yeah. So I quote from the internet that SQL injection is application vulnerability that occurs when untrusted data is inserted in a SQL query without any sanitation or escaping. So this is how it looks. If this particular line of code invoke with name FFF, the resulting query will be like this one. But if it's set to double quote, single quote, or one equal one, single quote one, the query will be like this one. The tag is the successful inclusion of an OR operator which help us return all the records from the database. And this is how SQL injection can be fixed. SQL injection is not possible when using the above because the first element of the array is a template and letters are parameters to the template. And for the second one, it can be used for the fixing SQL injection as well. So maybe I can show a little bit demo of SQL injection itself. So regardless to the programming language, I clone this sample project from GitLab. It's coming from Jason. Jason something. Thank you for him to make these simple apps. So it's basically, let's pretend it's an online store when we can search for something like water. Maybe we can type juice. What if we put, it will dump the database, the username, and the hash password of it. Yeah, I hope this kind of demo can give us more visibility to understand what is the SQL injection. Let's get back to the slide. Where's the slide? This one. And then what else besides SQL injection? There's a lot of which we can consider as a security issue. And for you guys, if you like deep more about the security issue, I would like to recommend you to visit the OAPS pages. So OAPS stands for Open Web Application Security Project. This is non-profit organization. Annually, they make 10 most critical web application security risk for each year. So let's say I take a sample. This is the top 10 security issue at the 2017. So because previously, we only talked about SQL injection. So OAPS itself categorized into several categories, like injection, broken authentication, and ETC. You can find it by yourselves at oasp.org. And how to make sure the code doesn't contain security coding bugs. So this is for the developers. So mostly of the developers, they don't think carefully about how to write in secure code. Is there any of you who has secure developer courses or secure development program? Nope. If that the case, I hope this topic will much easier help for you guys. And so how the developer can able to identify whether he writing the code securely or not. There is a tool called static application security testing, or commonly known as SAS. So SAS is designed to analyze source code and our compiled version of code to help find security flows. In this case, the flows which caused by coding bugs. Since today, we are in Ruby conference. So there is a tool for SAS. It's open source tool. It's called with Brackman. Has anyone used Brackman before? So I think it should be enough. Because everyone knows Brackman, so there's no point for me to continue the speech, right? So it's open source tools. We can use it for scan the Ruby on Rails application. It's statically analyzed race application code to find security issues. We can find it on the github.com slash president beef slash Brackman. So Brackman installation itself, it's not quite hard. We can install using Ruby games, using Bandler, or we can use the Docker version as well. And for the compatibility, Brackman should work with any version Rails from 2.3 to 5. something. Brackman can analyze code write-on with Ruby 1.8 syntax and newer, but requires at least Ruby 2.3.0 to run. And Brackman itself has several capability to find security issue on the coding box level, such as cross-site scripting, SQL injection, command injection, cross-site request forgery, and many more. And Brackman output format, it can be CSV, JSON, text, HTML. We can choose our preferences, whichever we want. And Brackman itself able to integrate with Jenkins. So if you guys using Jenkins for the CI CD orchestrator, we can integrate the Brackman into the CI CD pipeline as well. Because the Brackman has its plug-in for Jenkins. And maybe I can show you a quick demo about the Brackman usage. So I clone some Rails project. It's called with Rails Goat. So it's kind of like vulnerable apps for a base on Ruby on Rails. This is the Rails Goat project looks like. And for using Brackman, we can simply using this command. It's simple as Brackman and the project route. And we can type enter. She'll be working. And the result will coming through. And for a better user interface, we can convert it to certain file, like I mentioned earlier, to HTML. Maybe it's better if we put it to HTML. So HTML. So this is how the Brackman report looks like in HTML format. So first, it tells us the application path, which already being scanned. This is the path. And it tell us the information about the Rails version and the Brackman itself version. And it tell us what is being checked from Brackman. So we can see it's checked for basic authentication, JSON parsing, SQL, XSS, and et cetera. So if you take a look on here, it's coming with the warning type. Warning type, it's represent how many security issues which found on the application. It's already categorized based on this category, common injection, cross-site request forgery, SQL injection, et cetera. So in the next table, it tell us about the class, dashboard controller, the method, and then the warning type, let's say, for the SQL injection. And it's coming with the code snippet as well. For this one, we take a look for the SQL injection. And it tell us exactly the line number which potentially contains SQL injection. And it's provide us the link for the further info. If we click on the link, it will direct to the Brackman site and will tell us what is SQL injection and give us the suggestion how to fix it. I think that's a quick demo for the Brackman. Maybe for the next, there's a bundler audit. This is used for checking for the authorization of games in game file log. We can clone from github.com, rubysec, bundler audit. So maybe I show you the quick demo. So the comment is as simple as that, like just bundler hypen audit. And the result will come. It will tell us the Rails God application is using Nocogiri dependencies using version 1.10.2. And it has security issues with the high critical. And it's already listed on the vulnerable database library with CVE 2019, 5477. And this is the URLs. And this is the title. So this is the solution to upgrade from this version to this version to avoid the security issue. And as simple as that. And then what else do we have? I think that's all that I have. So just for the disclaimer, I'm not the ruby expert. I'm not kind of like hands-on guys on the ruby. But I'm still learning ruby. So if you have any question regarding to this, please don't hit me with the hard one. I think thank you. OK, we do have one question here, which is what are the most common security issues you see which Rails doesn't cover by default and Breakman doesn't catch? So basically, for figure out the security issues, it depends on the capability of tools itself. If we are talking about the security issue which caused by the coding bugs. So coding bugs mean we write the code. So if the question, what is the security issue which is not exposed by the hardest one, to be honest, I can't answer that kind of question, but my answer is it depends back on the tool. So as long as the tool has the capability, has the set rule to identify the certain bugs, it always catch by the tools. OK. OK, so I hope that's my help. Yeah, that's great. Thank you very much. Yeah, because the tools itself working based on the rule set. So for the example, for the school injection or another common security issue, the tools has the capability because it registered the rule set to identify those vulnerabilities. So if one day the new vulnerability occurs and for the current tools, it's not update the rule set, it's high probability that the tools could not identify the issue by itself. OK, thank you very much to Harley-Davidson. Thank you.


Hi everyone, Joe for jazzy's casebreaks.com coming at you with a fun one here 2023 gold rush shoebox collection Baseball edition three box break random team break number 130 total hits 10 hits per box. So who knows what we can get out of here Pretty nice checklist here Also, if we run into an issue where they're in their college gear or their minor league here I don't know how common that's gonna be but If that happens we'll default to the checklist first There's no checklist. I won't go by or no checklist rules active players will go to the team. They're currently on Retired players will go to the team. They played for the longest per baseball reference dot-com big. Thanks to this group for making it happen Congrats to the people who won their spot in that fanatics under wraps break Thanks to the people who bought their spots straight up All 30 teams are in let's roll it and randomized names and teams one out of five six times each one two three four five and six Steve it down to Jason Five one out of five six times for the teams one two three four five and sixth and final time after six we've got the Angels all the way down to the fight and fill is the Phillies Training my Dodgers by Four runs here All right Steve you got the Angels Chris with the Yankees Scott with the Mets Jim with the Royals Ed with the Nationals Nicole with the Orioles Steve with the Brewers Tristan with the Tigers Adam with my Dodgers Barry with the Astros Michael with the White Sox Jonathan with the Guardian Steve Locke with the Padres Austin with the Reds Eugene with the A's and Cardinals Jared with the Rangers Victor with the Marlins Scott with the Giants Jason with the Cubs Jason Kersh with the Rockies Aaron with the Red Sox Tristan with the Diamondbacks David with the Twins Tristan with the Rays Steve with the Pirates Joseph K with the Braves Chris with the Blue Jays and Jason K with the Philadelphia Phillies All right, let's sort by team And we're gonna pause the video When we come back we're gonna see if there's any trades then we'll have the break stick around BRB All right, welcome back everybody. No deals were done. So the list is Is now officially printed out remains the same. Thanks everyone for watching for breaking with us for joining the action There it is All right, and here are the three shoeboxes that we're gonna do so each box Has ten items per box. That's good design on here Got the gold rush sticker right there. So let's pop this baby open. Let's see what we got There's all sorts of fun stuff in here All right, let's just go from the top And we've got a cut auto Well, not a cut auto, but just a check Joe Sewell authentic autograph Often they use stuff like this for cut autos cut those out. What do you think he was getting he he wrote a check? Kids that are watching. This is how people used to pay for things They would know debit cards they didn't have cash you could write a check Joe Sewell wrote a check for three dollars and thirteen cents. What does that say the the pure The pure oil company it's paying for gas Yeah, that's what they did in the 60s Jilo. They wrote three dollar checks. I think Joe Sewell Played most of his career with the the tribe Or was he a Red Sox? No, he he spent 11 seasons with the Cleveland Indians at the time from 1920 to 1930 and then finished his career for the few seasons with the Yankees and So that'll be for Jonathan Cleveland. This is for you Next is Next is a baseball Ed Howard Ed Howard the fourth Ed Howard I think is a cub right? Yeah, he's that 21 year old cubs prospect That's gonna go to Jason Parker in the cubbies It's a nice penmanship there for for a kid. You got Guy let's show you how about the Jersey? Juan Soto That's pretty nice nice, so obviously Nationals colors It's like it's a it's a custom Jersey. So not officially licensed Got the JSA card and the sticker and all that stuff there. That'll be for Ed P in the Nats How about some cards? And you got Nolan aeronado 9 out of 15 Cardinals edition going to Eugene Nice Steve Carlton 24 out 25 dual relic and autograph from Sterling. That's for Jason K in the Phillies Nice Rhino 22 out of 50 Jason with the Cubs And a couple graded cards. Oh Frank Thomas Authentic Auto trust me. She'll like this too White Sox Michael and a PSA 10 Fernando de Dice Jr. On his rookie card Steve Locke with the Padres So what else we got a mini helmet and a photo here for the pirates We got Matt Lava lavalier got a price on there, too Anyone any Pirates fans who remember this guy he played for the Pirates back in The late 80s early 90s and gold lover in 1987. He's proud of that All right, that's gonna be for Steve Locke and Pittsburgh and the photo is For Cleveland That is a Shane Bieber, right? Yeah, I think so. Yeah, that's Shane Bieber's auto And all the Beckett stuff right there as well. So that's that's for Cleveland. That would be for Jonathan All right, that's the first box. I could put all this up back in that same box Go with the next box. What's happening here? A lot of fun stuff in here All right, let's start on what's going on. Let's just start off Right from the top Spencer Howard Rated Rookie Otto gold vinyl one of one Phillies Jason K All aboard the big hit express My Dodgers up eight one now on the Phillies bottom of the fourth This is cool. Sweet Lou Brock Post card it's kind of hard angle here, but authentic autograph Post card. There's this facsimile autograph. There's his real autograph in blue People who send postcards back in the day It's a thing That goes to Eugene and the Cardinals More cards here. How about an Alec Bum 15 out of 28? Nice RPA for the Phillies Jason K last one mojo Here's another Rhino 49 out of 50 Jason Parker and the Cubs And we got a CJ Abrams Nice German 10 auto grade German 10 Another one for Steve Locke another Padre for Steve Locke Nice. All right, we got Adrienne Beltrae nice He's got some great penmanship Number stamp on the ball Pretty cool all of your authentication right there Beltrae, of course played for the Rangers the longest Now you played for eight seasons with the Rangers Started his career with my Dodgers and that will be for Jared Jared P and the Rangers Nice Adrian Beltrae what future Hall of Famer. I would I would think I think there was it was kind of weird I think he hit 48 home runs in his last season with the Dodgers I think there may have been some PED stuff there Then went to Seattle and did okay, but then just really just grind it out at just a bunch of consistent Years with the Rangers ended up with 3,000 hits a lifetime 286 Average almost for almost 500 home runs 477. So I'll eventually get in there accusations. Yeah, accusations This is nice Can anyone anyone guess that autograph? Adrian Gonzalez Does some studio analysts work with the Dodgers here and there? I think post career, I think they Think he kind of co-owned Co-manage and played for a Mexican League team Yep, Joe Pizzles got it. It's Wander Franco for the Rays There's all the proper stickers right there and that is for Tristan in Tampa Bay. Nice. All right. Oh a couple jerseys in here number 33 for the A's Jose Conceco and That will be for Eugene and the Oakland A's and We got a Mets to Grom Jersey. There's his autograph right there. There's the JSA stuff Oh and Joe Pizzles saying that that wander is a signature debut Yeah with the SD Yeah, it's a signature debut right there Yeah, this item certifies a signature debut and is our is our considered opinion that the item is genuine signed before the last game With the signers official rookie season All right, so and Mets Scott with the Mets, I'm not sure I think Scott some it's got a met fan Yeah, Scott was saying I got my met, but I don't know if he wants that to grow All right, and final one here is Yep We know this guy Julio Rodriguez For Eugene all the Beckett stuff right there as well. Oh You'll take it's Scott will take them. All right. All right in the third and final gold rush shoe box Man, this was a lot of fun. We don't have any more move this, but This was a lot of fun It'll take me a second to just Take me a second to put all this stuff away, but So fun break just one Dodger says Adam. We're not done yet. Adam Play to the whistle man. We've got ten more hits to go All right, we have We have another Adrian Beltray There's all the stickers right there. This is his name stamp right there. So another one for the Rangers another baseball Well, we haven't seen this guy yet This autograph might be too clean Anyone tell this autograph right here might feel just read it and figure it out. Where did he play for the longest? Play from 1975 to 1992 18 year career Man, even in his age 33 season about four seasons before he hung it up Was a top 12 in MVP voting A couple of seasons before that was top three in MVP voting all-star silver slugger lifetime 267 340 home runs 1800 hits over 1200 walks it's kind of a That 31 year old season where he was a cardinal Had 136 walks 139 strikeouts an OPS of 1.055 That many walks and strikeouts wow Gilo can't recursive really no one's getting this it's jack Clark Played for the Giants the longest 10 years with the Giants Now jack the Ripper All right, that's for scott and the Giants Yeah, jack Clark Jack the Ripper It's his nickname It's killing baseballs. I guess all right some cards. Yeah, they're they're floating on top of another jersey here Then we got another check Christian Yelich eight out of ten quad relic an autograph For the Brewers that'll be for steve Solid shot jersey an autograph chipper jones chip off the old block out of 25 That'll be for joseph k and the braves I got a louise robert autograph michael gabbi and the white socks And then we've got a authentic autograph Ken griffey jr Mariners that's going to be for Eugene the signed check is Is nice stan the man musial What did he pay for january of 1987 59 50? Maybe something to maybe he owed some friends some money j and k harding Maybe for dinner or something like that Here i'll write you a check for dinner James and and kathy harding Nice stan musial Cardinals did he play for the Cardinals his entire career? I want to say yes, right? One of those one of those rare ones Yeah, play for the Cardinals his entire 22 year career hauling Oh, not harding hauling All right, so that's what it cost to haul something Back in the late 80s Do you think j and k hauling is still around victor? Yeah, they are j and k hauling On 33 28 west adam street st. Charles, Missouri. I'm assuming that's st louis area No, yelp reviews for them all right Here's a nice one for the braves akunia jr ronald akunia jr hit a bomb today For his 220 someoth homer homer looks like a Not a official jersey, but it's an official autograph for the braves. That'll be for joseph k You got another jersey Everyone remember this guy We've got a photo on the bottom here Some of you old school guys may remember old rick demsey World series mvp. I think it's with the dodgers for a little bit, too Orioles that's going to be for nicole And the last one here is is Is that blue vita blue and mike tower also saying vita blue. I'm looking up his autograph. Yes Vita blue oakland a's that's for eugine That was a lot of fun ladies and gentlemen that was 2023 gold rush shoe box shoe box collection Baseball edition three box random team rake number one. I'm joe for jaspy's case rakes.com. Thanks for watching Thanks for breaking with us and I'll see you next time for the next one. Bye. Bye