%I #5 Mar 16 2015 02:00:34
%S 1,2,15,308,13399,1019106,119698377,20039968920,4527610159068,
%T 1326616296092984,489092182592254708,221537815033845709776,
%U 120928125204565597029220,78286897353506845258973144,59305342759674536454338570652,51970719684035315747385128783808
%N Expansion of exp( Sum_{n >= 1} A210672(n)*x^n/n ).
%C It appears that this sequence is integer valued.
%C The o.g.f. A(x) = 1 + 2*x + 15*x^2 + 308*x^3 + ... for this sequence is such that 1 + x*d/dx( log(A(x) ) is the o.g.f. for A210672.
%C This sequence is the particular case m = 2 of the following general conjecture.
%C Let m be an integer and consider the sequence u(n) defined by the recurrence u(n) = m*Sum_{k = 0..n-1} binomial(2*n,2*k) *u(k) with the initial condition u(0) = 1. Then the expansion of exp( Sum_{n >= 1} u(n)*x^n/n ) has integer coefficients.
%C For cases see A255926(m = -3), A255882(m = -2), A255881(m = -1), A255928(m = 1) and A255930(m = 3).
%C Note that u(n), as a polynomial in the variable m, is the n-th row generating polynomial of A241171.
%F O.g.f.: exp(2*x + 26*x^2/2 + 842*x^3/3 + 50906*x^4/4 + ...) = 1 + 2*x + 15*x^2 + 308*x^3 + 13399*x^4 + ....
%F a(0) = 1 and a(n) = 1/n*Sum_{k = 0..n-1} A210672(n-k)*a(k) for n >= 1.
%p #A255929
%p A210672 := proc (n) option remember; if n = 0 then 1 else 2*add(binomial(2*n, 2*k)*A210672(k), k = 0 .. n-1) end if; end proc:
%p A255929 := proc (n) option remember; if n = 0 then 1 else add(A210672(n-k)*A255929(k), k = 0 .. n-1)/n end if; end proc:
%p seq(A255929(n), n = 0 .. 15);
%Y A210672, A241171, A255926(m = -3), A255882(m = -2), A255881(m = -1), A255928(m = 1), A255930(m = 3).
%K nonn,easy
%O 0,2
%A _Peter Bala_, Mar 11 2015