unexpected EIS hit
Vladeta
vladeta at eunet.yu
Sun Sep 1 05:15:08 CEST 2002
Sum_{i>= 1} x^i/(1-x^i)^(k+1) is g.f. for inverse Moebius transform of
binomial(n+k-1,k) = 1/k!* Sum_{i=1..k} abs(stirling1(k,i))*sigma[i](n).
Regards,
Vladeta
__________________
----- Original Message -----
From: "wouter meeussen" <wouter.meeussen at pandora.be>
To: "Seqfan (E-mail)" <seqfan at ext.jussieu.fr>
Sent: Saturday, August 31, 2002 5:59 PM
Subject: unexpected EIS hit
>
> define f[a[1..N],n] as Sum(d|n, a[1]a[d]+a[2]a[d-1]+..+a[d]a[1] ),N>=n
>
> then f[ {1,2,3,4,..},n]=
> {1,5,11,25,36,71,85,145,176,260,287,455,456,649,726
> =ID Number: A059358
> =Coefficients in expansion of Sum_{n >= 1} x^n/(1-x^n)^4
>
> I don't see the 'why'. Who does?
>
> the moebius transform of A059358 is
> A000292 (Tetrahedral (or pyramidal) numbers: C(n+3,3),
> Also the convolution of the natural numbers with themselves )
> ..and this makes sense (Moebius inversion formula).
>
> But why Sum x^n/(1-x^n)^4 ???
> Any combinatorial significance?
>
> in Mathematica:
> f[li:{__Integer},n_Integer]:=Fold[#1+ (Reverse at Take[li,#2]. Take[li,#2])
> &,0,Divisors[n]];
> Table[f[Range[48],k],{k,48}]
>
>
> Wouter Meeussen
> wouter.meeussen at pandora.be
>
>
>
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