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Search: id:A089677
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| A089677 |
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Exponential convolution of A000670(n), with A000670(0)=0, with the sequence of all ones alternating in sign. |
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+0 4
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| 0, 1, 1, 7, 37, 271, 2341, 23647, 272917, 3543631, 51123781, 811316287, 14045783797, 263429174191, 5320671485221, 115141595488927, 2657827340990677, 65185383514567951, 1692767331628422661
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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Stirling transform of A005212(n)=[1,0,6,0,120,0,5040,...] is a(n)=[1,1,7,37,271,...]. - Michael Somos Mar 04 2004
Occurs also as first column of a matrix-inversion occuring in a sum-of-like-powers problem. Consider the problem for any fixed natural number m>2 of finding solutions to sum(k=1,n,k^m) = (k+1)^m. Erdos conjectured that there are no solutions for n,m>2. Let D be the matrix of differences of D[m,n] := sum(k=1,n,k^m) - (k+1)^m. Then the generating functions for the rows of this matrix D constitute a set of polynomials in n (for varying n along columns) and the m-th polynomial defining the m-th row. Let GF_D be the matrix of the coefficients of this set of polynomials. Then the present sequence is the (unsigned) second column of GF_D^-1. - Gottfried Helms, Apr 01 2007
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LINKS
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Gottfried Helms, Discussion of a problem concerning summing of like powers
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FORMULA
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E.g.f.: (exp(x)-1)/(exp(x)*(2-exp(x))). a(n)=Sum(Binomial(n, k)(-1)^(n-k)Sum(i! Stirling2(k, i), i=1, ..k), k=0, .., n).
a(n) = (A000670(n)-(-1)^n)/2. - Vladeta Jovovic (vladeta(AT)Eunet.yu), Jan 17 2005
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MATHEMATICA
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Table[Sum[Binomial[n, k](-1)^(n-k)Sum[i! StirlingS2[k, i], {i, 1, k}], {k, 0, n}], {n, 0, 20}]
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PROGRAM
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(PARI) a(n)=if(n<0, 0, n!*polcoeff(subst(y/(1-y^2), y, exp(x+x*O(x^n))-1), n))
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CROSSREFS
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Cf. A052841.
Sequence in context: A102760 A096965 A100309 this_sequence A075996 A093168 A097493
Adjacent sequences: A089674 A089675 A089676 this_sequence A089678 A089679 A089680
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KEYWORD
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easy,nonn
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AUTHOR
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Mario Catalani (mario.catalani(AT)unito.it), Jan 03 2004
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