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

%I A228767 #18 Oct 10 2013 11:51:32
%S A228767 -2,-9,-45,-231,-1161,-5643,-26637,-122895,-557073,-2490387,-11010069,
%T A228767 -48234519,-209715225,-905969691,-3892314141,-16642998303,
%U A228767 -70866960417,-300647710755,-1271310319653,-5360119185447,-22539988369449,-94557999988779,-395824185999405
%N A228767 Second bisection of the inverse binomial transform of the rational sequence with e.g.f. (x/2)*(exp(-x)+1)/(exp(x)-1).
%C A228767 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, ...
%C A228767 Its first bisection is constituted of fractional numbers, with denominators A176591, whereas this bisection is constituted of integers only.
%C A228767 It appears that a(1) = -2 and a(n) = -1 * A005408(n-1) * A087289(n-2) for n>1.
%F A228767 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]
%F A228767 Conjecture: a(n) = (8+4^n)*(1-2*n)/8 for n>1, a(1)=-2. [_Bruno Berselli_, Sep 03 2013]
%o A228767 (PARI) fr(n) = if (n==0, 1, (-1)^n*(subst(bernpol(n), x, 1) + subst(bernpol(n), x, 2))/2);
%o A228767 ibtfr(n) = sum(k = 0, n, (-1)^(n-k)*binomial(n, k) * fr(k));
%o A228767 lista(nn) = {forstep(n=1, nn, 2, print1(ibtfr(n), ", "););} \\ _Michel Marcus_, Sep 03 2013
%K A228767 sign
%O A228767 1,1
%A A228767 _Michel Marcus_, following a suggestion of _Paul Curtz_, Sep 03 2013