OFFSET
1,1
COMMENTS
For any m>0, sum(k=0,n,4^(m*k)*B(2*k)*C(2*n+1,2*k)) is always an integer. sum(k=0,n,4^k*B(2*k)*C(2*n+1,2*k)) = 2n+1.
FORMULA
It seems that a(n) is asymptotic to (n!)^2*w*z^n where z = 1.63....and w = ? [There is a missing factor sqrt(n), z = 16/Pi^2 = 1.6211389382774... - Vaclav Kotesovec, Feb 15 2019]
a(n) ~ (n!)^2 * 2^(4*n + 3) * sqrt(n) / Pi^(2*n + 3/2). - Vaclav Kotesovec, Feb 15 2019
MATHEMATICA
Table[(-1)^(n+1)*Sum[16^k*BernoulliB[2*k]*Binomial[2*n + 1, 2*k], {k, 0, n}], {n, 1, 20}] (* Vaclav Kotesovec, Feb 15 2019 *)
PROG
(PARI) a(n)=(-1)^(n+1)*sum(k=0, n, bernfrac(2*k)*binomial(2*n+1, 2*k)*16^k)
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Jan 10 2003
STATUS
approved