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a(n) = n^5 - (n-1)^5 + (n-2)^5 - ... +(-1)^n*0^5.
5

%I #26 Jun 13 2015 00:50:28

%S 0,1,31,212,812,2313,5463,11344,21424,37625,62375,98676,150156,221137,

%T 316687,442688,605888,813969,1075599,1400500,1799500,2284601,2869031,

%U 3567312,4395312,5370313,6511063,7837844,9372524,11138625,13161375

%N a(n) = n^5 - (n-1)^5 + (n-2)^5 - ... +(-1)^n*0^5.

%C 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

%H Harry J. Smith, <a href="/A062393/b062393.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (5,-9,5,5,-9,5,-1).

%F a(n) = (2*n^5 + 5*n^4 - 5*n^2 + 1 - (-1)^n)/4 = n^5 - a(n-1).

%F G.f.: x*(x^4 + 26*x^3 + 66*x^2 + 26*x + 1)/((x-1)^6*(x+1)). - _Colin Barker_, Sep 19 2012

%F 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

%p a := n -> (1-(-1)^n+n^2*(n^2*(2*n+5)-5))/4; # _Peter Luschny_, Jul 12 2009

%t k=0;lst={k};Do[k=n^5-k;AppendTo[lst, k], {n, 1, 5!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Dec 11 2008 *)

%t 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 *)

%o (PARI) { a=0; for (n=0, 1000, write("b062393.txt", n, " ", a=n^5 - a) ) } \\ _Harry J. Smith_, Aug 07 2009

%Y Cf. A000539, A000584. A062392 for 4th powers, A152725 for 6th powers.

%K nonn,easy

%O 0,3

%A _Henry Bottomley_, Jun 21 2001