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A299424 G.f. A(x) satisfies: 1 = Sum_{n>=0} binomial((n+1)*(n+2), n) / ((n+1)*(n+2)/2) * x^n / A(x)^((n+1)*(n+2)/2).

%I A299424 #13 Feb 14 2018 09:23:58
%S A299424 1,2,3,16,214,4268,110520,3493432,130094524,5576660656,270624533544,
%T A299424 14679297273336,880950418422088,57999873220558048,4159207521351394736,
%U A299424 322857188993783461264,26982279001217007103104,2416289398989507579797760,230882321991151748812644944,23451581238859769517764695488,2523663354227008958133677737296
%N A299424 G.f. A(x) satisfies: 1 = Sum_{n>=0} binomial((n+1)*(n+2), n) / ((n+1)*(n+2)/2) * x^n / A(x)^((n+1)*(n+2)/2).
%C A299424 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 A299424 Paul D. Hanna, <a href="/A299424/b299424.txt">Table of n, a(n) for n = 0..200</a>
%e A299424 G.g.: A(x) = 1 + 2*x + 3*x^2 + 16*x^3 + 214*x^4 + 4268*x^5 + 110520*x^6 + ...
%e A299424 such that
%e A299424 1 = 1/A(x) + 2*x/A(x)^3 + 11*x^2/A(x)^6 + 114*x^3/A(x)^10 + 1827*x^4/A(x)^15 + 40508*x^5/A(x)^21 + 1159587*x^6/A(x)^28 + ...
%o A299424 (PARI) {a(n) = my(A=[1]); for(i=1, n, A = Vec(sum(m=0, #A, binomial((m+1)*(m+2), m)/((m+1)*(m+2)/2) * x^m/Ser(A)^((m+1)*(m+2)/2-1) ))); A[n+1]}
%o A299424 for(n=0, 30, print1(a(n), ", "))
%Y A299424 Cf. A299434.
%K A299424 nonn
%O A299424 0,2
%A A299424 _Paul D. Hanna_, Feb 13 2018