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G.f. A(x) satisfies: 1 = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n / A(x)^((n+1)^2).
2

%I #23 Feb 17 2018 23:47:03

%S 1,1,1,6,77,1451,35730,1082481,38913817,1619979291,76724619427,

%T 4077896446598,240566693095072,15609120639706252,1105414601508493001,

%U 84881459931003622118,7026832554316541379141,624014794413319426058889,59184228450018585954486975,5971678912361406721742217080,638782082648832471805820934833,72213308562202419209594988387550,8603323896642095980014195130664418

%N G.f. A(x) satisfies: 1 = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n / A(x)^((n+1)^2).

%C Compare to: 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.

%H Paul D. Hanna, <a href="/A299434/b299434.txt">Table of n, a(n) for n = 0..200</a>

%e G.f.: A(x) = 1 + x + x^2 + 6*x^3 + 77*x^4 + 1451*x^5 + 35730*x^6 + 1082481*x^7 + 38913817*x^8 + 1619979291*x^9 + 76724619427*x^10 +...

%e such that

%e 1 = 1/A(x) + C(4,1)/4*x/A(x)^4 + C(9,2)/9*x^2/A(x)^9 + C(16,3)/16*x^3/A(x)^16 + C(25,4)/25*x^4/A(x)^25 + C(36,5)/36*x^5/A(x)^36 + C(49,6)/49*x^6/A(x)^49 + ...

%e more explicitly,

%e 1 = 1/A(x) + x/A(x)^4 + 4*x^2/A(x)^9 + 35*x^3/A(x)^16 + 506*x^4/A(x)^25 + 10472*x^5/A(x)^36 + 285384*x^6/A(x)^49 + ... + A143669(n)*x^n/A(x)^((n+1)^2) + ...

%o (PARI) {a(n) = my(A=[1]); for(i=1,n, A = Vec(sum(n=0,#A,binomial((n+1)^2,n)/(n+1)^2 * x^n/Ser(A)^((n+1)^2-1) )));G=Ser(A);A[n+1]}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A298692, A143669.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Feb 13 2018