0,3

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 (pevnev(AT)juno.com), Jan 09 2008

Contribution from Peter Luschny (peter(AT)luschny.de), Jul 12 2009: (Start)

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^(-4)(P(3,1)-(-1)^k P(3,2k+1))|. (End)

Problem 913 of the Spring 1997 issue of the Pi Mu Epsilon Journal.

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.

Skidmore College Problem Group, Solution to Problem #913 from the Pi Mu Epsilon Journal

(1/8)[ -1 + (-1)^n - 6*(-1)^n*n^2 - 4*(-1)^n*n^3 ]. - Henry Bottomley (se16(AT)btinternet.com), Nov 13 2000

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 (se16(AT)btinternet.com), Nov 13 2000

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 (pevnev(AT)juno.com), Apr 26 2008

{-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

a := n -> ((2*n+3)*n^2-(n mod 2))/4; [From Peter Luschny (peter(AT)luschny.de), Jul 12 2009]

k=0; lst={k}; Do[k=n^3-k; AppendTo[lst, k], {n, 1, 5!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 11 2008]

Sequence in context: A100752 A162024 A143058 this_sequence A159222 A100206 A007044

Adjacent sequences: A011931 A011932 A011933 this_sequence A011935 A011936 A011937

nonn,easy

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

More terms from Henry Bottomley (se16(AT)btinternet.com), Nov 13 2000