%I #53 Nov 03 2024 09:33:53

%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,2*k+1)). Thus we get expression a(k) = |2^(-4)*(P(3,1)-(-1)^k P(3,2*k+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*(2*n-1)/4). - _Henry Bottomley_, Nov 13 2000

%F G.f.: x*(1 + 4*x + x^2)/(1 - 3*x + 2*x^2 + 2*x^3 - 3*x^4 + x^5). - _Alexander R. Povolotsky_, Apr 26 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 Table[(4*n^3 -6*n^2 +1-(-1)^n)/8, {n,0,100}] (* _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 *)

%o (Magma) [((2*n+3)*n^2 - (n mod 2))/4: n in [0..100]]; // _G. C. Greubel_, Nov 03 2024

%o (SageMath) [((2*n+3)*n^2 - (n%2))//4 for n in range(101)] # _G. C. Greubel_, Nov 03 2024

%Y Cf. A015238, A032528, A153641.

%K nonn,easy,changed

%O 0,3

%A David Penney (david(AT)math.uga.edu)

%E More terms from _Henry Bottomley_, Nov 13 2000