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

%I A249348 #24 Mar 12 2020 09:44:32
%S A249348 0,1,28,1378,111628,13507003,2282683528,513603793828,148431496416328,
%T A249348 53583770206294453,23630442660975853828,12500504167656226675078,
%U A249348 7812815104785141671923828,5695542211388368278832470703,4789950999777617722498107861328
%N A249348 a(n) = (A001147(n+1)^2-1)/8, where A001147(n+1) = 3*5*...*(2n+1).
%C A249348 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.)
%C A249348 This is a subsequence of the centered 9-gonal numbers A060544, which are a subsequence of the triangular numbers A000217.
%H A249348 S. Klein, <a href="http://lnkd.in/bKzkYcS">A neat infinite sum ...</a>, "Number Theory" group on LinkedIn, Oct. 2014.
%F A249348 (-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
%p A249348 A249348 := proc(n)
%p A249348     (doublefactorial(2*n+1)^2-1)/8 ;
%p A249348 end proc:
%p A249348 seq(A249348(n),n=0..20) ;
%o A249348 (PARI) a(n)=(prod(k=1,n,2*k+1)^2-1)/8
%K A249348 nonn
%O A249348 0,3
%A A249348 _M. F. Hasler_, Oct 26 2014
%E A249348 a(11)/a(12) corrected by _Georg Fischer_, Mar 12 2020