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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?2011-08-18
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    @Jadmrial Are there no typos? As written, the value of $a_0$ doesn't influence others $a_n$.2011-08-18
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    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.2011-08-18
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    Sorry, I've made a typo. I've fixed it in the first post. The formula should have been $a_n=\left(\frac{-1}{\lambda^b}\right)^n\frac{\Gamma(nb+1)}{n!}$. I also made an error in simplifying a term- simple exponent error (sorry, I was a little tired when I posted the query).2011-08-18
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    Peter Bala: Those values for the sequence aren't correct. They should be: $1,-1,2,-\frac{16}{3},\frac{352}{21},\frac{-2048}{35},\frac{2142208}{9765},\cdots$2011-08-18
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    The values I gave were for the question in its original form with $a_{n-1}$ in the sum, rather than $a_{n-i}$ as you have now. What do you calculate the values of $b$ and $\lambda$ to be to make your formula for $a_{n}$ fit the sequence?2011-08-18
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    Plugging the formula into the recurrence, using the integral definition for $\Gamma$, simplifying with the binomial theorem, and then identifying back with the original formula, yields $$\Gamma(nb+1)=\frac{1}{1-2^{1-n}}\int_0^\infty\int_0^\infty\left[(x^b+y^b)^n-x^{bn}-y^{bn}\right]e^{-(x+y)}dxdy.$$ Not sure what to do from there.2011-08-18
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    Peter: I'm pretty sure it's always been $a_{n-i}$. Values: $b\approx 1.64677...$, $\lambda\approx 1.48149$ anon: Thanks.2011-08-18
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    Using the generating function$$g(x)=\sum_{k=1}^\infty a_k x^k$$and considering the recurrence relation above, we get that$$g(x)-2g(x/2) = g(x)^2$$2011-08-18
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    @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.2011-08-23
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    @Jadmrial: From [a tabulation of the first few values](http://www.wolframalpha.com/input/?i=table+of+%28%28-1%2F1.48149+%5E1.64677%29%5En+gamma%281.64677+n+%2B+1%29+%2F+n%21%29), it seems that you didn't even check the equality for $n=1$?!2011-08-23

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