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A298692 G.f. A(x) satisfies: 1 = Sum_{n>=0} binomial( n*(n+1), n)/(n+1) * x^n / A(x)^( (n+1)^2 ).

%I A298692 #13 Oct 17 2020 04:18:43
%S A298692 1,1,2,15,213,4485,123566,4171778,166069875,7602292250,393220294258,
%T A298692 22679300697606,1443478702575162,100529312696403699,
%U A298692 7606562231567559478,621526322941129712986,54553240678513466719077,5120001583257750960650134,511729676123794537164792892,54270040973557127212080028474,6087267497390906756985330494931
%N A298692 G.f. A(x) satisfies: 1 = Sum_{n>=0} binomial( n*(n+1), n)/(n+1) * x^n / A(x)^( (n+1)^2 ).
%C A298692 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 A298692 Paul D. Hanna, <a href="/A298692/b298692.txt">Table of n, a(n) for n = 0..100</a>
%F A298692 a(n) ~ c * d^n * n! / n^2, where d = -4 / (LambertW(-2*exp(-2)) * (2 + LambertW(-2*exp(-2)))) = 6.17655460948348035823168... and c = 0.16989702... - _Vaclav Kotesovec_, Oct 17 2020
%e A298692 G.f.: A(x) = 1 + x + 2*x^2 + 15*x^3 + 213*x^4 + 4485*x^5 + 123566*x^6 + 4171778*x^7 + 166069875*x^8 + 7602292250*x^9 + 393220294258*x^10 + 22679300697606*x^11 + 1443478702575162*x^12 + 100529312696403699*x^13 + 7606562231567559478*x^14 + 621526322941129712986*x^15 + ...
%e A298692 such that
%e A298692 1 = 1/A(x) + C(2,1)/2*x/A(x)^4 + C(6,2)/3*x^2/A(x)^9 + C(12,3)/4*x^3/A(x)^16 + C(20,4)/5*x^4/A(x)^25 + C(30,5)/6*x^5/A(x)^36 + C(42,6)/7*x^6/A(x)^49 + C(56,7)/8*x^7/A(x)^64 + ...
%e A298692 more explicitly,
%e A298692 1 = 1/A(x) + x/A(x)^4 + 5*x^2/A(x)^9 + 55*x^3/A(x)^16 + 969*x^4/A(x)^25 + 23751*x^5/A(x)^36 + 749398*x^6/A(x)^49 + 28989675*x^7/A(x)^64 + ... + A135861(n)*x^n/A(x)^((n+1)^2) + ...
%t A298692 terms = 21; A[_] = 1; Do[A[x_] = A[x] - 1 + Sum[Binomial[n*(n+1), n]/(n+1)*x^n/A[x]^((n + 1)^2) + O[x]^(terms), {n, 0, k}], {k, terms}]; CoefficientList[A[x], x] (* _Jean-François Alcover_, Feb 06 2018 *)
%o A298692 (PARI) {a(n) = my(A=[1]); for(i=1,n, A = Vec(sum(m=0,#A,binomial(m*(m+1),m)/(m+1) * x^m/Ser(A)^((m+1)^2-1) ))); A[n+1]}
%o A298692 for(n=0,20,print1(a(n),", "))
%Y A298692 Cf. A135861.
%K A298692 nonn
%O A298692 0,3
%A A298692 _Paul D. Hanna_, Feb 03 2018