A228767 Second bisection of the inverse binomial transform of the rational sequence with e.g.f. (x/2)*(exp(-x)+1)/(exp(x)-1).

-2, -9, -45, -231, -1161, -5643, -26637, -122895, -557073, -2490387, -11010069, -48234519, -209715225, -905969691, -3892314141, -16642998303, -70866960417, -300647710755, -1271310319653, -5360119185447, -22539988369449, -94557999988779, -395824185999405

Offset: 1

Author

Michel Marcus, following a suggestion of Paul Curtz, Sep 03 2013

Keywords

• sign

Comments

The sequence to be transformed is A176328/A176591, its inverse binomial transform begins: 1, -2, 25/6, -9, 599/30, -45, 4285/42, -231, 15599/30, -1161, 169625/66, -5643, 33578309/2730, ...

Its first bisection is constituted of fractional numbers, with denominators A176591, whereas this bisection is constituted of integers only.

It appears that a(1) = -2 and a(n) = -1 * A005408(n-1) * A087289(n-2) for n>1.

Programs

PARI

fr(n) = if (n==0, 1, (-1)^n*(subst(bernpol(n), x, 1) + subst(bernpol(n), x, 2))/2);
ibtfr(n) = sum(k = 0, n, (-1)^(n-k)*binomial(n, k) * fr(k));
lista(nn) = {forstep(n=1, nn, 2, print1(ibtfr(n), ", "););} \\ Michel Marcus, Sep 03 2013

Formula

Conjecture: G.f. -x*(2-11*x+21*x^2-2*x^3+8*x^4)/((1-x)^2*(1-4*x)^2). [Bruno Berselli, Sep 03 2013]

Conjecture: a(n) = (8+4^n)*(1-2*n)/8 for n>1, a(1)=-2. [Bruno Berselli, Sep 03 2013]