A255930 - OEIS

%I #4 Mar 16 2015 02:00:43

%S 1,3,33,991,63060,7018860,1206748720,295775068680,97835325011235,

%T 41970842737399345,22655642596496388759,15025240474194493147857,

%U 12008582230377080862401692,11382727559611560650861409564,12625404970864692720119281536900,16199644066580777034289339157904220

%N Expansion of exp( Sum_{n >= 1} A210674(n)*x^n/n ).

%C It appears that this sequence is integer valued.

%C The o.g.f. A(x) = 1 + 3*x + 33*x^2 + 991*x^3 + ... for this sequence is such that 1 + x*d/dx( log(A(x) ) is the o.g.f. for A210674.

%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 A255926(m = -3), A255882(m = -2), A255881(m = -1), A255928 (m = 1) and A255929(m = 2).

%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(3*x + 57*x^2/2 + 2703*x^3/3 + 239277*x^4/4 + ...) = 1 + 3*x + 33*x^2 + 991*x^3 + 63060*x^4 + ....

%F a(0) = 1 and a(n) = 1/n*Sum_{k = 0..n-1} A210674(n-k)*a(k) for n >= 1.

%p #A255930

%p A210674 := proc (n) option remember; if n = 0 then 1 else 3*add(binomial(2*n, 2*k)*A210674(k), k = 0 .. n-1) end if; end proc:

%p A255930 := proc (n) option remember; if n = 0 then 1 else add(A210674(n-k)*A255930(k), k = 0 .. n-1)/n end if; end proc:

%p seq(A255930(n), n = 0 .. 15);

%Y Cf. A210674, A241171, A255926(m = -3), A255882(m = -2), A255881(m = -1), A255928(m = 1), A255929(m = 2).

%K nonn,easy

%O 0,2

%A _Peter Bala_, Mar 11 2015