A157308 G.f. A(x) satisfies the condition that both A(x) and F(x) = A(x*F(x)) = g.f. of A155585 have zeros for every other coefficient after initial terms; g.f. of dual sequence A157309 satisfies the same condition.

1, 1, -1, 0, 3, 0, -38, 0, 947, 0, -37394, 0, 2120190, 0, -162980012, 0, 16330173251, 0, -2070201641498, 0, 324240251016266, 0, -61525045423103316, 0, 13913915097436287598, 0, -3698477457114061621492, 0

Offset: 0

Author

Paul D. Hanna, Mar 11 2009

Keywords

• sign

Comments

After initial 2 terms, reversing signs yields A157310.

Conjecture: a(m) == 1 (mod 2) iff m is a power of 2 or m=0. [Paul D. Hanna, Mar 17 2009]

Examples

			G.f.: A(x) = 1 + x - x^2 + 3*x^4 - 38*x^6 + 947*x^8 - 37394*x^10 +-...
RELATED FUNCTIONS.
If F(x) = A(x*F(x)) then F(x) = o.g.f. of A155585:
A155585 = [1,1,0,-2,0,16,0,-272,0,7936,0,-353792,0,...];
...
If G(x) = A(x*G(x))/(1+x) then G(x) = o.g.f. of A122045:
A122045 = [1,0,-1,0,5,0,-61,0,1385,0,-50521,0,2702765,0,...];
...
		

Crossrefs

Cf. A157309, A157310, A157304, A157305, A155585, A122045 (Euler numbers).

Cf. A158119. [Paul D. Hanna, Mar 17 2009]

Programs

Mathematica

terms = 28;
F[x_] = Sum[n! x^n/Product[(1 + 2k x), {k, 1, n}], {n, 0, terms+1}] + O[x]^(terms+1);
A[x_] = x/InverseSeries[x F[x]];
CoefficientList[A[x], x][[1 ;; terms]] (* Jean-François Alcover, Jul 26 2018 *)

PARI

{a(n)=local(A=[1, 1]); for(i=1, n, if(#A%2==0, A=concat(A, 0);); if(#A%2==1, A=concat(A, t); A[ #A]=-subst(Vec(x/serreverse(x*Ser(A)))[ #A], t, 0))); Vec(x/serreverse(x*Ser(A)))[n+1]}

Formula

Let F(x) = o.g.f. of A155585, then o.g.f. A(x) satisfies:

A(x) = x/serreverse(x*F(x));

A(x) = 2x + F( -x/(A(x) - 2x) );

A(x) = F(x/A(x));

F(x) = A(x*F(x));

where A155585 is defined by e.g.f. exp(x)/cosh(x).

...

Let G(x) = o.g.f. of A122045, then o.g.f. A(x) satisfies:

A(x) = x + x/serreverse(x*G(x));

A(x) = x + G( x/(A(x) - x) );

G(x) = A(x*G(x))/(1+x);

where A122045 is the Euler numbers.

...

O.g.f.: A(x) = 2*(1+x) - H(x) where H(x) = g.f. of A157310.