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A295764 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1) * x^n/A(x)^n.

%I A295764 #26 Apr 20 2024 03:49:32
%S A295764 1,2,8,84,1522,38940,1278880,51136512,2407921070,130495143060,
%T A295764 8002598818976,548083639109204,41481408076217480,3439242379130298624,
%U A295764 310088460754461374608,30213233395705566500248,3164041937175571224372118,354458174755809132668847604,42301583754989091491716706272,5358139267082652649000120167456
%N A295764 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1) * x^n/A(x)^n.
%H A295764 Paul D. Hanna, <a href="/A295764/b295764.txt">Table of n, a(n) for n = 0..300</a>
%F A295764 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = binomial((n+1)^2,n) for n>=0.
%F A295764 a(n) ~ c * exp(n) * n^(n - 3/2), where c = exp(3/2 - 2*exp(-2)) / sqrt(2*Pi) = 1.3639581816297942453935... - _Vaclav Kotesovec_, Aug 11 2021, updated Apr 20 2024
%e A295764 G.f.: A(x) = 1 + 2*x + 8*x^2 + 84*x^3 + 1522*x^4 + 38940*x^5 + 1278880*x^6 + 51136512*x^7 + 2407921070*x^8 + 130495143060*x^9 + 8002598818976*x^10 +...
%e A295764 such that
%e A295764 A(x) = 1 + 2*(x/A(x)) + 12*(x/A(x))^2 + 140*(x/A(x))^3 + 2530*(x/A(x))^4 + 62832*(x/A(x))^5 + 1997688*(x/A(x))^6 +...+ binomial((n+1)^2,n)/(n+1)*(x/A(x))^n +...
%e A295764 The table of coefficients of x^k in A(x)^(n+1) begins:
%e A295764   [1, 2, 8, 84, 1522, 38940, 1278880, 51136512, 2407921070, ...];
%e A295764   [1, 4, 20, 200, 3444, 85312, 2744928, 108267280, 5049708672, ...];
%e A295764   [1, 6, 36, 356, 5862, 140508, 4424984, 172064160, 7946443350, ...];
%e A295764   [1, 8, 56, 560, 8888, 206176, 6350112, 243284064, 11121338640, ...];
%e A295764   [1, 10, 80, 820, 12650, 284252, 8556240, 322780800, 14599990830, ...];
%e A295764   [1, 12, 108, 1144, 17292, 376992, 11084864, 411518448, 18410660208, ...];
%e A295764   [1, 14, 140, 1540, 22974, 487004, 13983816, 510586400, 22584587382, ...];
%e A295764   [1, 16, 176, 2016, 29872, 617280, 17308096, 621216192, 27156348512, ...];
%e A295764   [1, 18, 216, 2580, 38178, 771228, 21120768, 744800256, 32164253550, ...]; ...
%e A295764 in which the main diagonal begins:
%e A295764   [1, 4, 36, 560, 12650, 376992, 13983816, ..., binomial((n+1)^2,n), ...],
%e A295764 thus [x^n] A(x)^(n+1) = [x^n] (1 + x)^((n+1)^2) for n>=0.
%t A295764 terms = 20; A[_] = 1; Do[A[x_] = Sum[Binomial[(n+1)^2, n]/(n+1)*x^n/A[x]^n, {n, 0, terms}] + O[x]^terms // Normal, terms];
%t A295764 CoefficientList[A[x], x] (* _Jean-François Alcover_, Jan 14 2018 *)
%o A295764 (PARI) {a(n) = my(A=[1]); for(m=1,n, A = concat(A,0); V = Vec( Ser(A)^(m+1) ); A[m+1] = (binomial((m+1)^2,m) - V[m+1])/(m+1);); A[n+1]}
%o A295764 for(n=0,20,print1(a(n),", "))
%Y A295764 Cf. A295765, A295763, A070780.
%K A295764 nonn
%O A295764 0,2
%A A295764 _Paul D. Hanna_, Jan 06 2018