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%I #8 Mar 16 2015 02:00:01
%S 1,-3,30,-802,45414,-4508190,692197470,-151610017950,44827810930305,
%T -17193060505570335,8298004578522898140,-4920774627129981351120,
%U 3516683319021255757053900,-2980761698101283167670391780,2956463734237276273792194346560,-3392220222832838757465019626175680
%N Expansion of exp( Sum_{n >= 1} A210676(n)*x^n/n ).
%C It appears that this sequence is integer valued.
%C The o.g.f. A(x) = 1 - 3*x + 30*x^2 - 802*x^3 + ... for this sequence is such that 1 + x*d/dx( log(A(x) ) is the o.g.f. for A210676.
%C This sequence is the particular case m = -3 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 A255882(m = -2), A255881(m = -1), A255928(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 polynomial of A241171.
%F O.g.f.: exp(-3*x + 51*x^2/2 - 2163*x^3/3 + 171231*x^4/4 + ...) = 1 - 3*x + 30*x^2 - 802*x^3 + 45414*x^4 - ....
%F a(0) = 1 and a(n) = 1/n*Sum_{k = 0..n-1} A210676(n-k)*a(k) for n >= 1.
%p #A255926
%p A210676 := proc (n) option remember; if n = 0 then 1 else -3*add(binomial(2*n, 2*k)*A210676(k), k = 0 .. n-1) end if; end proc:
%p A255926 := proc (n) option remember; if n = 0 then 1 else add(A210676(n-k)*A255926(k), k = 0 .. n-1)/n end if; end proc:
%p seq(A255926(n), n = 0 .. 16);
%Y Cf. A210676, A241171, A255882(m = -2), A255881(m = -1), A255928(m = 1), A255929(m = 2), A255930(m = 3).
%K sign,easy
%O 0,2
%A _Peter Bala_, Mar 11 2015
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