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.)
LINKS
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
PROG
(PARI) a(n)=(prod(k=1, n, 2*k+1)^2-1)/8
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Oct 26 2014
EXTENSIONS
a(11)/a(12) corrected by Georg Fischer, Mar 12 2020
STATUS
approved