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A229128 a(n) = (1/(n+1)) * Sum_{k=0..n} A185072(n-k)*A185072(k).

%I A229128 #10 Sep 21 2013 09:24:39
%S A229128 1,1,-1,2,-8,10,-372,-6014,-194140,-6851310,-286013372,-13629729990,
%T A229128 -731279661488,-43618952847894,-2863512195607780,-205184821861288518,
%U A229128 -15936259396425487780,-1333664198543385424414,-119648378626870160202060,-11456079894214769377662158
%N A229128 a(n) = (1/(n+1)) * Sum_{k=0..n} A185072(n-k)*A185072(k).
%C A229128 Does this sequence consist entirely of integers?
%C A229128 The g.f. G(x) of A185072 satisfies [x^(n+1)] G(x)^(n^2+1) = 0 for n>0.
%H A229128 Paul D. Hanna, <a href="/A229128/b229128.txt">Table of n, a(n) for n = 0..200</a>
%e A229128 G.f.: A(x) = 1 + x - x^2 + 2*x^3 - 8*x^4 + 10*x^5 - 372*x^6 - 6014*x^7 +...
%e A229128 The g.f. of A185072 begins:
%e A229128 G(x) = 1 + x - 2*x^2 + 6*x^3 - 28*x^4 + 70*x^5 - 1446*x^6 - 22302*x^7 +...
%e A229128 G(x)^2 = 1 + 2*x - 3*x^2 + 8*x^3 - 40*x^4 + 60*x^5 - 2604*x^6 - 48112*x^7 +...
%e A229128 where the coefficient of x^n in G(x)^2 appears to be divisible by (n+1) for n>=0.
%o A229128 (PARI) {a(n)=local(A=[1, 1]); for(i=1, n, A=concat(A, 0); A[#A]=-Vec((1/x*serreverse(x/Ser(A)^(#A-1)))^(1/(#A-1)))[#A]); Vec(Ser(A)^2)[n+1]/(n+1)}
%o A229128 for(n=0, 30, print1(a(n), ", "))
%Y A229128 Cf. A185072.
%K A229128 sign
%O A229128 0,4
%A A229128 _Paul D. Hanna_, Sep 14 2013