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%I #12 Oct 01 2016 00:27:45
%S 1,0,0,1,0,0,30,0,0,10921,0,0,6308995,0,0
%N Inverse binomial transform of A277041.
%F Let G(x) be the g.f. of A277042, then g.f. A(x) satisfies:
%F (1) G(x*A(x)) = (1+x)*A(x).
%F (2) A(x/(G(x) - x)) = G(x) - x.
%F (3) A(x) = (1/x)*Series_Reversion(x/(G(x) - x)).
%F (4) G(x) = x + x/Series_Reversion(x*A(x)).
%e G.f.: A(x) = 1 + x^3 + 30*x^6 + 10921*x^9 + 6308995*x^12 +...
%e such that the binomial transform forms the g.f. of A277041:
%e A(x/(1-x))/(1-x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 11*x^5 + 51*x^6 + 246*x^7 + 897*x^8 + 13526*x^9 + 115631*x^10 + 614681*x^11 + 8739556*x^12 + 89877217*x^13 + 596072842*x^14 +...+ A277041(n)*x^n +...
%e Also, A(x/(G(x) - x)) = G(x) - x where G(x) = g.f. of A277042 where
%e G(x) = 1 + x + x^3 + 27*x^6 + 10666*x^9 + 6174792*x^12 +...+ A277042(n)*x^n +...
%Y Cf. A202582, A277040, A277041, A277042.
%K nonn,more
%O 0,7
%A _Paul D. Hanna_, Sep 25 2016
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