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
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).
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
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