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%I #44 Jun 14 2023 06:08:38
%S 0,1,7,20,44,81,135,208,304,425,575,756,972,1225,1519,1856,2240,2673,
%T 3159,3700,4300,4961,5687,6480,7344,8281,9295,10388,11564,12825,14175,
%U 15616,17152,18785,20519,22356,24300,26353,28519,30800,33200,35721,38367,41140
%N a(n) = |1^3 - 2^3 + 3^3 - 4^3 + ... + (-1)^(n+1)*n^3|.
%C From the formula a(n) = n^3 - a(n-1) it follows that a(n-1) + a(n) = n^3. Thus the sum of two consecutive terms (call them the "former" and "latter" terms) is a cube of the index of the "latter" term. - _Alexander R. Povolotsky_, Jan 09 2008
%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 we get expression a(k) = |2^(-4)(P(3,1)-(-1)^k P(3,2k+1))|. - _Peter Luschny_, Jul 12 2009
%C a(n) is the number of (w,x,y} having all terms in {0,...,n} and w<floor((x+y)/2). Also, the number of (w,x,y} having all terms in {0,...,n} and w>=floor((x+y)/2). [_Clark Kimberling_, Jun 02 2012]
%D Eldon Hansen's _A Table of Series and Products_ (Prentice-Hall, 1975) gives the sum in Formula 6.2.2 in terms of Euler polynomials.
%H Stanislav Sykora, <a href="/A011934/b011934.txt">Table of n, a(n) for n = 0..1000</a>
%H Kenneth B. Davenport, <a href="http://www.pme-math.org/journal/issues/PMEJ.Vol.10.No.6.pdf">Problem 913</a>, Pi Mu Epsilon Journal, Vol. 10, No. 6, Spring 1997, p. 492.
%H Skidmore College Problem Group, <a href="http://www.skidmore.edu/academics/mcs/pme913.htm">Solution to Problem #913 from the Pi Mu Epsilon Journal</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3, -2, -2, 3, -1).
%F a(n) = |(1/8)*(-1 + (-1)^n - 6*(-1)^n*n^2 - 4*(-1)^n*n^3)|. - _Henry Bottomley_, Nov 13 2000
%F a(n) = n^3-a(n-1) = a(n-1)+A032528(n) = ceiling(A015238(n+1)/4) = ceiling((n+1)^2*(2n-1)/4). - _Henry Bottomley_, Nov 13 2000
%F G.f.: (x^3 + 4*x^2 + x)/(x^5 - 3*x^4 + 2*x^3 + 2*x^2 - 3*x + 1). - _Alexander R. Povolotsky_, Apr 26 2008
%F {-a(n)-a(n+1)+n^3+3*n^2+3*n+1, a(0)=0, a(1)=1, a(2)=7, a(3)=20}. - Robert Israel, May 14 2008
%F a(n) = Sum_{k=1..n} floor((2*n+1)*k/2). - _Wesley Ivan Hurt_, Apr 01 2017
%p a := n -> ((2*n+3)*n^2-(n mod 2))/4; # _Peter Luschny_, Jul 12 2009
%t k=0;lst={k};Do[k=n^3-k;AppendTo[lst, k], {n, 1, 5!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Dec 11 2008 *)
%t Table[(4 n^3 - 6 n^2 - (-1)^n + 1)/8, {n, 1, 100}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 28 2011 *)
%t Abs[Accumulate[Times@@@Partition[Riffle[Range[0,50]^3,{1,-1},{1,-1,2}],2]]] (* _Harvey P. Dale_, May 20 2019 *)
%K nonn,easy
%O 0,3
%A David Penney (david(AT)math.uga.edu)
%E More terms from _Henry Bottomley_, Nov 13 2000
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