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%I #5 Mar 16 2015 02:02:42
%S 1,1,4,44,1025,41693,2617128,234091692,28251572652,4421489003700,
%T 870650503128708,210629395976568828,61405707768736724472,
%U 21231253444779700476672,8589776776743377081599500,4020181599664131540547091076,2155088041310451318611119556661
%N Expansion of exp( Sum_{n >= 1} A094088(n)*x^n/n ).
%C It appears that this sequence is integer valued.
%C The o.g.f. A(x) = 1 + x + 4*x^2 + 44*x^3 + ... for this sequence is such that 1 + x*d/dx( log(A(x) ) is the o.g.f. for A094088.
%C This sequence is the particular case m = 1 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), A255929(m = 2) 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(x + 7*x^2/2 + 121*x^3/3 + 3907*x^4/4 + ...) = 1 + x + 4*x^2 + 44*x^3 + 1025*x^4 + ....
%F a(0) = 1 and a(n) = 1/n*Sum_{k = 0..n-1} A094088(n-k)*a(k) for n >= 1.
%p #A255928
%p A094088 := proc (n) option remember; if n = 0 then 1 else add(binomial(2*n, 2*k)*A094088(k), k = 0 .. n-1) end if; end proc:
%p A255928 := proc (n) option remember; if n = 0 then 1 else add(A094088(n-k)*A255928(k), k = 0 .. n-1)/n end if; end proc:
%p seq(A255928(n), n = 0 .. 16);
%Y Cf. A094088, A241171, A255926(m = -3), A255882(m = -2), A255881(m = -1), A255929(m = 2), A255930(m = 3).
%K nonn,easy
%O 0,3
%A _Peter Bala_, Mar 11 2015
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