|
%I #2 Mar 30 2012 18:37:16
%S 1,4,16,384,17856,13492992,11507268608,160888878129152,
%T 2306486569154275328,537309590223329223155712,
%U 126767209261235580163634135040,483356141899716284828508078471905280
%N G.f.: A(x) = exp( Sum_{n>=1} 4^[(n^2+1)/2]*x^n/n ), a power series in x with integer coefficients.
%C It appears that g.f. exp( Sum_{n>=1} m^[(n^2+1)/2]*x^n/n ) forms a power series in x with integer coefficients for any positive integer m.
%F a(n) = (1/n)*Sum_{k=1..n} 4^floor((k^2+1)/2) * a(n-k) for n>0, with a(0)=1.
%e G.f.: A(x) = 1 + 4*x + 16*x^2 + 384*x^3 + 17856*x^4 + 13492992*x^5 +...
%e log(A(x)) = 4*x + 4^2*x^2/2 + 4^5*x^3/3 + 4^8*x^4/4 + 4^13*x^5/5 + 4^18*x^6/6 +...
%o (PARI) {a(n)=polcoeff(exp(sum(k=1, n, 4^floor((k^2+1)/2)*x^k/k)+x*O(x^n)), n)}
%Y Cf. A156335, A156336, A155207, A155200.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Feb 10 2009
| 