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Let $\Sigma$ be Radó's Busy Beaver function, and let $\Delta[\Sigma]$ denote the forward difference of $\Sigma$, such that $\Delta[\Sigma] \ (n) = \Sigma(n+1) - \Sigma(n)$ for all $n \in \mathbb{N}$. Define $\Delta^0[\Sigma] = \Sigma$, and $\Delta^{k+1}[\Sigma] = \Delta[\Delta^{k}[\Sigma]]$, for all $k \in \mathbb{N}$.

Question: Is it the case that for all $k \in \mathbb{N}$, the function $\Delta^k [\Sigma]$ eventually dominates every computable function $f: \mathbb{N} \rightarrow \mathbb{N}$?

(I believe this is the case, but do not know how to prove it. Any pointers or sources on proving this -- or disproving it, if I'm wrong?)

Here's a difference table showing all the known values for these sequences:

     n:  0      1      2      3      4      5     ...
     Σ:  0      1      4      6      13    ≥4098  ...
    ΔΣ:  1      3      2      7     ≥4085  ...
   ΔΔΣ:  2     -1      5     ≥4078  ...
  ΔΔΔΣ: -3      6     ≥4073  ...
 ΔΔΔΔΣ:  9     ≥4067  ...
ΔΔΔΔΔΣ: ≥4058  ...

Known sources: "A note on Busy Beavers and other creatures" (1996), by Ben-Amram, Julstrom, and Zwick, contains the conjecture that for any computable function $f$, there exists a constant $N_f$ such that $\forall n \gt N_f, \Sigma(n+1) > f(\Sigma(n))$. In support of this they prove the weaker result that for any computable function $f$, $\Sigma(n+1) > f(\Sigma(n))$ for infinitely many values of $n$. (Considering $f(x) = 2x$, for example, the conjecture implies that $\Delta[\Sigma]$ eventually dominates every computable function.)

Motivation: In the "Busy Beaver game" class of Turing machines, which eventually halt after starting with a blank tape, the "score" of a machine is the number of $1$s remaining on the tape after halting. A variant of Kolmogorov complexity $C: \mathbb{N} \rightarrow \mathbb{N}$ defines $C(n)$ to be the least $k$ such that $n$ is the score of some $k$-state machine. (cf "Computability and Logic" by Boolos, Burgess & Jeffrey, 2007, p229.) I believe I can show that if the first difference $\Delta[\Sigma]$ eventually dominates every computable function, then this $C$ is not monotonically nondecreasing, i.e., there exist $m \lt n$ such that $C(m) \gt C(n)$.

NB: Due to receiving no satisfactory answer here, I've posted the simplest case of this question on cstheory.SE.

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Have you tried posting this question on MathOverflow? –  Quinn Culver Feb 21 '12 at 15:41

2 Answers 2

For the eventual conclusion "there exist $m<n$ such that $C(m)>C(n)$" it seems to be easier just to prove that for some $k$ there exists a $k$-state machine with a score higher than the number of possible $k$-state machines. Then by the pigeonhole principle there has to be a number less than this that is reached only by a machine with $>k$ states.

Depending on the details, there are at most about $(6k)^{2k}$ different $k$-state machines, and this is easily a computable function. Therefore if add some initial states that preload a large enough $k$ (which can be done in $O(\log k)$ states) we will eventually reach a machine with a score that is high enough relative to its size.

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Good points. Which approach is easier depends on how hard it is to prove the domination result for the first difference of $\Sigma$ (and maybe that's harder than I thought, assuming it's true!). But given that result, and the fact that it's easy to show $C(\Sigma(k-1) + 1) = C(\Sigma(k)) = k$, the non-monotonicity conclusion follows readily by considering the number of $k$-state machines as a computable function of $k$ that's eventually dominated by $\Sigma(k) - \Sigma(k-1)$. –  r.e.s. Nov 14 '11 at 16:45
Hmmm ... I upvoted your reply although it's not an answer to my question -- not realizing that this removes the question from the "unanswered" list. I really would like to have a good answer re the $k$th difference of $\Sigma$. I think it would be a noteworthy result even to hold for only $k = 1$ (but I suspect it either holds for no positive $k$, or else it holds for all positive $k$). –  r.e.s. Nov 15 '11 at 19:03

Use http://mathworld.wolfram.com/NewtonsForwardDifferenceFormula.html to get $D^k [z] (n)>f(k,n)z(n+k)$ for some computable $f(k,n)$ and every strictly increasing positive function $z$. Thus if $g(n)>D^k [z](n)$ for some computable $g$, then there is a computable bound on any $z(n+k)$. Contradiction.

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I am still puzzled. Where do you get the first sentence from? It might also help if you specify exactly what $f(k,n)$ is. –  Srivatsan Nov 14 '11 at 13:36
(FYI: I've switched from $D$ to the standard $\Delta$ notation for the forward difference.) –  r.e.s. Nov 16 '11 at 4:04

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