A249348 a(n) = (A001147(n+1)^2-1)/8, where A001147(n+1) = 3*5*...*(2n+1).

0, 1, 28, 1378, 111628, 13507003, 2282683528, 513603793828, 148431496416328, 53583770206294453, 23630442660975853828, 12500504167656226675078, 7812815104785141671923828, 5695542211388368278832470703, 4789950999777617722498107861328

Offset

0,3

Comments

These are the numerators of the partial sums S(n) = Sum_{k=1..n} A000217(k)/A001147(k+1)^2 before simplification, i.e., a(n) = S(n)*A001147(n+1)^2, where A000217(n) = n(n+1)/2. The series S(n) has sum 1/8, actually S(n) = 1/8 - 1/(8*A001147(n+1)^2). (Similarly, Sum_{n=1..oo} A249354(n)/A007559(n+1)^3 = 1/9, where A249354(n) = 3n^3+3n^2+n.)

This is a subsequence of the centered 9-gonal numbers A060544, which are a subsequence of the triangular numbers A000217.

Links

Table of n, a(n) for n=0..14.

S. Klein, A neat infinite sum ..., "Number Theory" group on LinkedIn, Oct. 2014.

Formula

(-n+1)*a(n) +2*n*(2*n^2-1)*a(n-1) -(n+1)*(-1+2*n)^2*a(n-2)=0. - R. J. Mathar, Oct 28 2014

Maple

A249348 := proc(n)
    (doublefactorial(2*n+1)^2-1)/8 ;
end proc:
seq(A249348(n), n=0..20) ;

Prog

(PARI) a(n)=(prod(k=1, n, 2*k+1)^2-1)/8

Crossrefs

Sequence in context: A091993 A213692 A118705 * A366302 A013926 A110696

Adjacent sequences: A249345 A249346 A249347 * A249349 A249350 A249351

Keyword

nonn

Author

M. F. Hasler, Oct 26 2014

Extensions

a(11)/a(12) corrected by Georg Fischer, Mar 12 2020

Status

approved