A059338 a(n) = Sum_{k=1..n} k^5 * binomial(n,k).

1, 34, 342, 2192, 11000, 47232, 181888, 646144, 2156544, 6848000, 20877824, 61526016, 176171008, 492126208, 1345536000, 3610247168, 9526771712, 24769069056, 63546720256, 161087488000, 403925630976, 1002841309184, 2467290939392, 6019866427392, 14575206400000, 35039249170432

Offset: 1

Author

Pat Costello (matcostello(AT)acs.eku.edu), Jan 26 2001

Keywords

• nonn
• easy

References

• Finding a closed form for the sum was Problem 541 in the Fall 2000 issue of The Pentagon (official journal of Kappa Mu Epsilon).

Links

• Harry J. Smith, Table of n, a(n) for n = 1..200
• Index entries for linear recurrences with constant coefficients, signature (12,-60,160,-240,192,-64).

Crossrefs

Binomial transform of A000584.

Programs

Maple

with(combinat): for n from 1 to 70 do printf(`%d,`,sum(k^5*binomial(n,k), k=1..n)) od:

Mathematica

Table[Sum[k^5*Binomial[n, k], {k,1,n}], {n,1,5}] (* or *) LinearRecurrence[{12, -60, 160, -240, 192, -64}, {1, 34, 342, 2192,
  11000, 47232}, 10] (* G. C. Greubel, Jan 07 2017 *)

PARI

a(n) = { sum(k=1, n, k^5*binomial(n, k)) } \\ Harry J. Smith, Jun 26 2009

PARI

Vec(x*(16*x^4-32*x^3-6*x^2+22*x+1)/(2*x-1)^6 + O(x^25)) \\ G. C. Greubel, Jan 07 2017

Formula

The closed form comes from starting with (1+x)^n and repeatedly differentiating and multiplying by x. After five differentiations, substitute x=1 and get a(n) = 2^(n-5)*n^2*(n^3+10n^2+15n-10).

G.f.: x*(16*x^4-32*x^3-6*x^2+22*x+1)/(2*x-1)^6. - Colin Barker, Sep 20 2012

Extensions

More terms from James Sellers, Jan 29 2001