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A375442 Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(4*n-3*k, k) * ([x^k] A(x)^n) for n >= 1.

%I A375442 #13 Sep 12 2024 16:05:32
%S A375442 1,1,4,58,2040,141471,17414299,3548205700,1133401180600,
%T A375442 542919997056985,375875913182499803,364645747772536556398,
%U A375442 482860134802496098766740,853352199545631748454314991,1973943714691108570803508891357,5875562740836479708875221002947800,22169669117423382437537944364347343144
%N A375442 Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(4*n-3*k, k) * ([x^k] A(x)^n) for n >= 1.
%C A375442 Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+k, 2*k) * ([x^k] C(x)^n) for n >= 1 is satisfied by the Catalan function C(x) = 1 + x*C(x)^2 (A000108), where coefficient [x^k] C(x)^n = binomial(n+2*k-1,k)*n/(n+k).
%C A375442 Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+2*k, 3*k) * ([x^k] D(x)^n) for n >= 1 is satisfied by the function D(x) = 1 + x*D(x)^3 (g.f. of A001764), where coefficient [x^k] D(x)^n = binomial(n+3*k-1, k)*n/(n+2*k).
%H A375442 Paul D. Hanna, <a href="/A375442/b375442.txt">Table of n, a(n) for n = 0..200</a>
%F A375442 From _Vaclav Kotesovec_, Sep 12 2024: (Start)
%F A375442 a(n) ~ c * d^n * n!^4 * n^alpha, where d = 4/Pi^4, alpha = 3*(1 + Pi*tanh(Pi/2))/2 = 5.821978559932543777976797..., c = 0.0936210705524085151687305...
%F A375442 Equivalently, a(n) ~ c * 2^(2*n) * n^(4*n + (7 + 3*Pi*tanh(Pi/2))/2) / (exp(4*n) * Pi^(4*n)), where c = 3.69601171983499372807019... (End)
%e A375442 G.f.: A(x) = 1 + x + 4*x^2 + 58*x^3 + 2040*x^4 + 141471*x^5 + 17414299*x^6 + 3548205700*x^7 + 1133401180600*x^8 + ...
%e A375442 RELATED TABLES.
%e A375442 The table of coefficients of x^k in A(x)^n begins:
%e A375442   n=1: [1, 1,  4,  58,  2040, 141471,  17414299, ...];
%e A375442   n=2: [1, 2,  9, 124,  4212, 287486,  35131224, ...];
%e A375442   n=3: [1, 3, 15, 199,  6528, 438267,  53158351, ...];
%e A375442   n=4: [1, 4, 22, 284,  9001, 594052,  71503584, ...];
%e A375442   n=5: [1, 5, 30, 380, 11645, 755096,  90175175, ...];
%e A375442   n=6: [1, 6, 39, 488, 14475, 921672, 109181745, ...];
%e A375442   ...
%e A375442 from which we may illustrate the defining property given by
%e A375442 0 = Sum_{k=0..n} (-1)^k * binomial(4*n-3*k, k) * ([x^k] A(x)^n).
%e A375442 Using the coefficients in the table above, we see that
%e A375442   n=1: 0 = 1*1 - 1*1;
%e A375442   n=2: 0 = 1*1 - 5*2 + 1*9;
%e A375442   n=3: 0 = 1*1 - 9*3 + 15*15 - 1*199;
%e A375442   n=4: 0 = 1*1 - 13*4 + 45*22 - 35*284 + 1*9001;
%e A375442   n=5: 0 = 1*1 - 17*5 + 91*30 - 165*380 + 70*11645 - 1*755096;
%e A375442   n=6: 0 = 1*1 - 21*6 + 153*39 - 455*488 + 495*14475 - 126*921672 + 1*109181745;
%e A375442   ...
%e A375442 The triangle T(n,k) = binomial(4*m-3*k, k) begins:
%e A375442   n=0: 1;
%e A375442   n=1: 1, 1;
%e A375442   n=2: 1, 5, 1;
%e A375442   n=3: 1, 9, 15, 1;
%e A375442   n=4: 1, 13, 45, 35, 1;
%e A375442   n=5: 1, 17, 91, 165, 70, 1;
%e A375442   n=6: 1, 21, 153, 455, 495, 126, 1;
%e A375442   ...
%o A375442 (PARI) {a(n) = my(A=[1],m); for(i=1, n, A=concat(A, 0); m=#A-1;
%o A375442 A[#A] = sum(k=0, m, (-1)^(m-k+1) * binomial(4*m-3*k, k) * polcoef(Ser(A)^m, k) )/m ); A[n+1]}
%o A375442 for(n=0, 20, print1(a(n), ", "))
%Y A375442 Cf. A375440, A375441, A375450, A375451.
%K A375442 nonn
%O A375442 0,3
%A A375442 _Paul D. Hanna_, Sep 12 2024