 Subtracting one from both sides we obtain

This is easily factored into

from which we obtain

and finally

Now comes the crucial step.We substitute this expression for "x" back into itself, recursively, to obtain

But now we can make the same recursive substitution again, and again, and again, pushing the unknown quantity "x" as far down and to the right as we please, and obtaining in the limit the infinite continued fraction

By applying the fundamental recurrence formulas we may easily compute the successive convergents of this continued fraction to be 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, ..., where each successive convergent is formed by taking the numerator plus the denominator of the preceding term as the denominator in the next term, then adding in the preceding denominator to form the new numerator.This sequence of denominators is a particular Lucas sequence known as the Pell numbers.