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A059338
a(n) = Sum_{k=1..n} k^5 * binomial(n,k).
5
1, 34, 342, 2192, 11000, 47232, 181888, 646144, 2156544, 6848000, 20877824, 61526016, 176171008, 492126208, 1345536000, 3610247168, 9526771712, 24769069056, 63546720256, 161087488000, 403925630976, 1002841309184
OFFSET
1,2
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
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
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 *)
PROG
(PARI) { for (n = 1, 200, write("b059338.txt", 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
CROSSREFS
Binomial transform of A000584.
Sequence in context: A281805 A034978 A251938 * A301954 A368719 A362953
KEYWORD
nonn,easy
AUTHOR
Pat Costello (matcostello(AT)acs.eku.edu), Jan 26 2001
EXTENSIONS
More terms from James A. Sellers, Jan 29 2001
STATUS
approved