[seqfan] Re: possible genfunc for A137341
Max Alekseyev
maxale at gmail.com
Wed Mar 11 23:06:12 CET 2009
From the recurrent formula for A137341, it follows that
A'(x) = exp(x) * A(x)
where A(x) = SUM[k=0..oo] A137341(k) * x^k / k!^2.
Solution to this differential equation is
A(x) = exp(exp(x))
PARI's serlaplace() simply removes k! from the denominator of the series.
Applied two times to exp(exp(x)), it gives series
SUM[k=0..oo] A137341(k) * x^k,
i.e., the ordinary generating function of A137341.
Regards,
Max
On Wed, Mar 11, 2009 at 12:58 AM, Joerg Arndt <arndt at jjj.de> wrote:
>
> http://www.research.att.com/~njas/sequences/A137341
>
> appears to be reproduced by (pari/gp):
>
> ? default(seriesprecision,16);
> ? Vec(serlaplace(serlaplace(exp(sum(k=1,16,1*x^k/(k!))))))
> [1, 1, 4, 30, 360, 6240, 146160, 4420080, 166924800, 7673823360,
> 420850080000, 27086342976000, 2018319704755200, 172142484203289600,
> 16642276683198566400, 1808459441303074560000,
> 219273812138054209536000]
>
> If anyone sees why this would be true, please
> add as a comment to the seq!
>
> How do we call such a generating function?
> EEGF ??
>
>
>
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