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

Olivier Gerard olivier.gerard at gmail.com
Sat Dec 31 12:32:11 CET 2011

```On Sat, Dec 31, 2011 at 03:19, Gottfried Helms <helms at uni-kassel.de> wrote:

> Dear seqfans -
>
>  I've discussed this problem already in another forum but could not
>  get a solution for the general case.
>
>  Consider the following formula for parameters r and c , (r,c >=0 )
>  which involves the Stirlingnumbers first kind and binomials:
>
> ------------------------------------------------------------------------
>
>
>           inf          s1(k+1, k-r+1)
> su(r,c) = sum  (-1)^k ---------------- * (-1)^c * (1+c)^k * binom(k+1,c+1)
>           k=a                k!
>
>           where   a = max(r,c)
>
> Hypothesis:    su(r,c) = 0  for all  r,c>=0
>
> ------------------------------------------------------------------------
>
> Empirically these sums are all tending to zero as the index k increases
> unboundedly, but I have no idea, how I could prove this in that
> generality for all indexes r and c.
> I used Mathematica (in its limited edition online at wolframalpha) for
> a couple of instances of r and c; and for r=0 and r=1 mathematica could
> even show, that the result is zero independently of the other parameter c.
> However, for bigger r the description of the stirling-numbers is too
> complicated - but note, that the matrix of Stirlingnumbers is evaluated
> along its subdiagonals!
>
>
Dear Gottfried,

Very interesting !

As Niels Abel would have said, adding a parameter can sometimes shed light
on
complicated sums.

If you try to sum with (1+c)^k  replaced by (1+x)^k  a symbolic calculator
can
give you an almost polynomial in x whose roots comprise c.
Here are the first ones:

Mathematica Code

Flatten[Table[{{r, c},
Sum[(-1)^k StirlingS1[k + 1, k - r + 1]/k! *(-1)^c*(1 + x)^k *
Binomial[k + 1, c + 1], {k, Max[c, r], Infinity}]}, {r, 1, 3}, {c, 1,
3}], 1]

(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.

Best regards,

Olivier

```