OFFSET
0,3
COMMENTS
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-6)(P(5,1)-(-1)^k P(5,2k+1))|. - Peter Luschny, Jul 12 2009
LINKS
Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-9,5,5,-9,5,-1).
FORMULA
a(n) = (2*n^5 + 5*n^4 - 5*n^2 + 1 - (-1)^n)/4 = n^5 - a(n-1).
G.f.: x*(x^4 + 26*x^3 + 66*x^2 + 26*x + 1)/((x-1)^6*(x+1)). - Colin Barker, Sep 19 2012
a(0)=0, a(1)=1, a(2)=31, a(3)=212, a(4)=812, a(5)=2313, a(6)=5463, a(n) = 5*a(n-1) - 9*a(n-2) + 5*a(n-3) + 5*a(n-4) - 9*a(n-5) + 5*a(n-6) - a(n-7). - Harvey P. Dale, Feb 01 2013
MAPLE
a := n -> (1-(-1)^n+n^2*(n^2*(2*n+5)-5))/4; # Peter Luschny, Jul 12 2009
MATHEMATICA
k=0; lst={k}; Do[k=n^5-k; AppendTo[lst, k], {n, 1, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
Table[Total[(Times@@@Partition[Riffle[Range[n, 0, -1], {1, -1}, {2, -1, 2}], 2])^5], {n, 0, 30}] (* or *) LinearRecurrence[ {5, -9, 5, 5, -9, 5, -1}, {0, 1, 31, 212, 812, 2313, 5463}, 40] (* Harvey P. Dale, Feb 01 2013 *)
PROG
(PARI) { a=0; for (n=0, 1000, write("b062393.txt", n, " ", a=n^5 - a) ) } \\ Harry J. Smith, Aug 07 2009
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Henry Bottomley, Jun 21 2001
STATUS
approved