# [seqfan] Re: A formula involving Stirling-numbers 1 kind, which seems to be zero for all parameters

Gottfried Helms helms at uni-kassel.de
Sat Dec 31 17:36:27 CET 2011

```Am 31.12.2011 12:32 schrieb Olivier Gerard:
> On Sat, Dec 31, 2011 at 03:19, Gottfried Helms <helms at uni-kassel.de> wrote:
>

>
> (r,c) sum value
>
> {{{1,1},1/4 E^(-1-x) (-1+x) (1+x) (-3-4 x+x^2)},
>
> {{1,2},1/12 E^(-1-x) (-2+x) (1+x)^2 (-1-6 x+x^2)},
>
> {{1,3},1/48 E^(-1-x) (-3+x) (1+x)^3 (3-8 x+x^2)},
>
> {{2,1},-(1/48) E^(-1-x) (-1+x) (1+x)^2 (43+77 x-35 x^2+3 x^3)},
>
> {{2,2},-(1/144) E^(-1-x) (-2+x) (1+x)^2 (-17+108 x+126 x^2-44 x^3+3 x^4)},
>
> {{2,3},-(1/576) E^(-1-x) (-3+x) (1+x)^3 (-89-48 x+246 x^2-56 x^3+3 x^4)},
>
> {{3,1},1/96 E^(-1-x) (-1+x) (1+x)^3 (-95-222 x+138 x^2-22 x^3+x^4)},
>
> {{3,2},1/288 E^(-1-x) (-2+x) (1+x)^3 (111-225 x-476 x^2+216 x^3-27
> x^4+x^5)},
>
> {{3,3},1/1152 (E^(-1-x) (-3+x) (1+x)^3 (187+808 x-341 x^2-904 x^3+313
> x^4-32 x^5+x^6))}}
>
>
> It shows that your formula is a cousin of the Dobinsky formula, but this
> times with StirlingS1, that it could perhaps be reformulated as a continous
> moment problem, that you have a family of non trivial polynomials, that
> probably can be expressed in terms of a finite binomial/stirling sum in
> terms of r and c, or perhaps prove that they always have a (x-c) factor.
>
Hi Olivier -

that was a great suggestion! I toyed around with something similar
by trying to formulate the sequence of stirlingnumbers in terms of
a polynomial earlier but didn't see the relevance; now this becomes clear.
I'll try to reproduce your above exposition using Pari/GP tomorrow or
the day next.
I think a *proof* of the x=c - root must use that polynomial expressions
which are a fixed property of the Stirling-constants.... Let's see...

Kind regards -

Gottfried

```