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%I #11 Sep 25 2016 19:51:27
%S 1,1,1,2,5,11,51,246,897,13526,115631,614681,8739556,89877217,
%T 596072842
%N Limit of the coefficient of x^(3^m + n) in B(x)^(n+1)/(n+1) as m grows, where B(x) = Sum_{k>=0} x^(3^k).
%C The expansion of A(x/(1+x))/(1+x) appears to be a power series in x^3.
%F a(n) = A277040(n)/(n+1).
%F Equals the binomial transform of A277043.
%e G.f.: A(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 +...
%e RELATED SERIES.
%e A(x/G(x)) = G(x) = x/Series_Reversion[x*A(x)], where
%e G(x) = 1 + x + x^3 + 27*x^6 + 10666*x^9 + 6174792*x^12 +...+ A277042(n)*x^n +...
%e and G(x) appears to continue with powers of x^3 only.
%e The inverse binomial transform forms the g.f. of A277043:
%e A(x/(1+x))/(1+x) = 1 + x^3 + 30*x^6 + 10921*x^9 + 6308995*x^12 +...+ A277043(n)*x^n +...
%e which also appears to continue with powers of x^3 only.
%o (PARI) { a(n) = my(m=n + ceil(log(n+3)/log(3)), B=sum(k=0, m, x^(3^k))); polcoeff((B+O(x^(3^m+n+1)))^(n+1)/(n+1), 3^m+n) }
%o for(n=0,10,print1(a(n),", "))
%Y Cf. A144691, A277040, A277042, A277043.
%K nonn,more
%O 0,4
%A _Paul D. Hanna_, Sep 25 2016
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