# Equivalence of two sequences

I'm having some trouble showing that two things I really want to be the same are in fact the same. I want to show that these two sequences are, in fact, the same thing:

$$a_0=1,a_1=-1, a_n=\sum_{i=1}^{n-1} \dfrac{a_ia_{n-i}}{1-2^{1-n}},\, n\geq 2$$

and

$$a_n=\left(\frac{-1}{\lambda^b}\right)^n\frac{\Gamma(nb+1)}{n!},$$ for some suitable $b$ and $\lambda$. (Actually, we know what each of these has to be just from the initial values $a_1,a_2$.)

I've tried some futzing about with the series expansions of the gamma function, and treating these as associated to some generating functions, but I'm running into walls. Some hints would be greatly appreciated.

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Have you tried replacing the $a_i$ in your purported recursion with the gamma function expression? – J. M. Aug 18 '11 at 10:10
@Jadmrial Are there no typos? As written, the value of $a_0$ doesn't influence others $a_n$. – Andrew Aug 18 '11 at 10:35
Using your recurrence, my computer produces the sequence $$1,-1,2,\frac{8}{3},\frac{704}{63},\frac{2106368}{11907},\frac{153888415121408}‌​{4395076119},...$$ The prime factorization of these numbers doesn't seem to show any pattern so I wouldn't expect them to be produced by a simple formula involving factorials and certainly not by the formula you suggest. By the way, putting the $a_{n-1}$ factor (which is independent of $i$) inside the summation looks rather odd. – Peter Bala Aug 18 '11 at 11:11
@Jadmrial: It's bad enough that your original post contained several typos; now you've twice added insult to injury by first claiming that Peter made a mistake and then even insisting that he was at fault after he explained the source of the problem. It's not a question of being "pretty sure"; you can just check the edit history to see that your original post had $n-1$. Please be more careful, both in order not to waste people's time, and also to avoid generating the impression that they made mistakes when they didn't. – joriki Aug 23 '11 at 14:40
@Jadmrial: From a tabulation of the first few values, it seems that you didn't even check the equality for $n=1$?! – joriki Aug 23 '11 at 15:24