A255926 Expansion of exp( Sum_{n >= 1} A210676(n)*x^n/n ).

1, -3, 30, -802, 45414, -4508190, 692197470, -151610017950, 44827810930305, -17193060505570335, 8298004578522898140, -4920774627129981351120, 3516683319021255757053900, -2980761698101283167670391780, 2956463734237276273792194346560, -3392220222832838757465019626175680

Offset: 0

Author

Peter Bala, Mar 11 2015

Keywords

• sign
• easy

Comments

It appears that this sequence is integer valued.

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.

This sequence is the particular case m = -3 of the following general conjecture.

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.

For cases see A255882(m = -2), A255881(m = -1), A255928(m = 1), A255929(m = 2) and A255930(m = 3).

Note that u(n), as a polynomial in the variable m, is the n-th row polynomial of A241171.

Crossrefs

Cf. A210676, A241171, A255882(m = -2), A255881(m = -1), A255928(m = 1), A255929(m = 2), A255930(m = 3).

Programs

Maple

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:
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:
seq(A255926(n), n = 0 .. 16);

Formula

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 - ....

a(0) = 1 and a(n) = (1/n)*Sum_{k = 0..n-1} A210676(n-k)*a(k) for n >= 1.