A115057 & A000914
zak seidov
zakseidov at yahoo.com
Fri Oct 12 20:11:15 CEST 2007
A115057 is A000914 with 0 prepended?
Zak
%N A000914 Stirling numbers of first kind: s(n+2,n).
%N A115057 Numbers of the form n(n^2-1)(3n+2)/24.
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>A115057 is A000914 with 0 prepended?
>Zak
>
>%N A000914 Stirling numbers of first kind: s(n+2,n).
>%N A115057 Numbers of the form n(n^2-1)(3n+2)/24.
According to the recursion for Stirling #s of the 1st Kind,
So, let a(n) = s(n+2,n).
Therefore, a(n) = (n+1)*s(n+1,n) + a(n-1).
Now, s(n+1,n) = n*s(n,n) + s(n,n-1). And s(n,n) = 1. And s(1,0)=0. So
Therefore, a(n) = n(n+1)^2/2 + a(n-1).
a(0) = 0, and n(n^2-1)(3n+2)/24 = 0 when n is 0.
And
n(n^2-1)(3n+2)/24 - (n-1)((n-1)^2-1)(3(n-1)+2)/24 = (n-1)n^2/2, which is
n(n+1)^2/2 with a shift of index.
So n(n^2-1)(3n+2)/24 = s(n+1,n-1).
Thanks,
Leroy Quet
>A115057 is A000914 with 0 prepended?
>Zak
>
>%N A000914 Stirling numbers of first kind: s(n+2,n).
>%N A115057 Numbers of the form n(n^2-1)(3n+2)/24.
Yes, they are the same except A000914 does not have the n=0 term.
Tony
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