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

1, 1, -1, 2, -8, 10, -372, -6014, -194140, -6851310, -286013372, -13629729990, -731279661488, -43618952847894, -2863512195607780, -205184821861288518, -15936259396425487780, -1333664198543385424414, -119648378626870160202060, -11456079894214769377662158

Offset: 0

Author

Paul D. Hanna, Sep 14 2013

Keywords

• sign

Comments

Does this sequence consist entirely of integers?

The g.f. G(x) of A185072 satisfies [x^(n+1)] G(x)^(n^2+1) = 0 for n>0.

Examples

			G.f.: A(x) = 1 + x - x^2 + 2*x^3 - 8*x^4 + 10*x^5 - 372*x^6 - 6014*x^7 +...
The g.f. of A185072 begins:
G(x) = 1 + x - 2*x^2 + 6*x^3 - 28*x^4 + 70*x^5 - 1446*x^6 - 22302*x^7 +...
G(x)^2 = 1 + 2*x - 3*x^2 + 8*x^3 - 40*x^4 + 60*x^5 - 2604*x^6 - 48112*x^7 +...
where the coefficient of x^n in G(x)^2 appears to be divisible by (n+1) for n>=0.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A185072.

Programs

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)}
for(n=0, 30, print1(a(n), ", "))