0,3

Number of edges in the join of two complete graphs of order n^2 and n, K_n^2 * K_n - Roberto E. Martinez II (remartin(AT)fas.harvard.edu), Jan 07 2002

a(n) is equal to the partial sums of A007588, stella octangula numbers: n(2n^2 - 1). - Jonathan Vos Post (jvospost3(AT)gmail.com), Mar 15 2006

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^(-5)(P(4,1)-(-1)^k P(4,2k+1))|. (End)

T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Harry J. Smith, Table of n, a(n) for n=0,...,1000

Milan Janjic, Two Enumerative Functions

a(n) = n(n+1)(n^2+n-1)/2 = n^4-a(n-1) = A000583(n)-a(n) = A000217(A028387(n-1)) = A000217(n)*A028387(n-1).

a(n) = SUM[i=0..n] A007588(i). a(n) = SUM[i=0..n] n*(2*n^2 - 1). a(n) = SUM[i=0..n] (1/6)*(12*n^3-6*n), n>0. - Jonathan Vos Post (jvospost3(AT)gmail.com), Mar 15 2006

a := n -> (2*n^2+n^3-1)*n/2; [From Peter Luschny (peter(AT)luschny.de), Jul 12 2009]

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

(PARI) { a=0; for (n=0, 1000, write("b062392.txt", n, " ", a=n^4 - a) ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Aug 07 2009]

Cf. A000538, A000583. A062393 provides the result for 5th powers, A011934 for cubes, A000217 for squares, A001057 (unsigned) for nonnegative integers, A000035 (offset) for 0th powers.

Cf. A007588.

Sequence in context: A027526 A033653 A088058 this_sequence A015876 A085474 A124893

Adjacent sequences: A062389 A062390 A062391 this_sequence A062393 A062394 A062395

nonn

Henry Bottomley (se16(AT)btinternet.com), Jun 21 2001