%I #11 Sep 12 2024 14:44:08
%S 1,1,5,65,1593,61953,3476813,265517441,26492540401,3349218907137,
%T 523572565927509,99215376614955457,22415450137196941993,
%U 5953820173628518544385,1837040977427662958973341,651657636773935012586716929,263375512326578915885862469601,120319850003020550647400856678401
%N Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(2*n, 2*k) * ([x^k] A(x)^n) for n >= 1.
%C 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).
%H Paul D. Hanna, <a href="/A375440/b375440.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) ~ c * 2^(4*n) * n^(2*n + 1/2) / (Pi^(2*n) * exp(2*n)), where c = 7.23682343848882192289996... - _Vaclav Kotesovec_, Sep 12 2024
%e G.f.: A(x) = 1 + x + 5*x^2 + 65*x^3 + 1593*x^4 + 61953*x^5 + 3476813*x^6 + 265517441*x^7 + 26492540401*x^8 + ...
%e RELATED TABLES.
%e The table of coefficients of x^k in A(x)^n begins:
%e n=1: [1, 1, 5, 65, 1593, 61953, 3476813, ...];
%e n=2: [1, 2, 11, 140, 3341, 127742, 7097687, ...];
%e n=3: [1, 3, 18, 226, 5259, 197637, 10869476, ...];
%e n=4: [1, 4, 26, 324, 7363, 271928, 14799444, ...];
%e n=5: [1, 5, 35, 435, 9670, 350926, 18895290, ...];
%e n=6: [1, 6, 45, 560, 12198, 434964, 23165174, ...];
%e ...
%e from which we may illustrate the defining property given by
%e 0 = Sum_{k=0..n} (-1)^k * binomial(2*n, 2*k) * ([x^k] A(x)^n).
%e Using the coefficients in the table above, we see that
%e n=1: 0 = 1*1 - 1*1;
%e n=2: 0 = 1*1 - 6*2 + 1*11;
%e n=3: 0 = 1*1 - 15*3 + 15*18 - 1*226;
%e n=4: 0 = 1*1 - 28*4 + 70*26 - 28*324 + 1*7363;
%e n=5: 0 = 1*1 - 45*5 + 210*35 - 210*435 + 45*9670 - 1*350926;
%e n=6: 0 = 1*1 - 66*6 + 495*45 - 924*560 + 495*12198 - 66*434964 + 1*23165174;
%e ...
%e The triangle A086645(n,k) = binomial(2*n, 2*k) begins:
%e n=0: 1;
%e n=1: 1, 1;
%e n=2: 1, 6, 1;
%e n=3: 1, 15, 15, 1;
%e n=4: 1, 28, 70, 28, 1;
%e n=5: 1, 45, 210, 210, 45, 1;
%e n=6: 1, 66, 495, 924, 495, 66, 1;
%e ...
%o (PARI) {a(n) = my(A=[1],m); for(i=1, n, A=concat(A, 0); m=#A-1;
%o A[m+1] = sum(k=0, m, (-1)^(m-k+1) * binomial(2*m, 2*k) * polcoef(Ser(A)^m, k) )/m ); A[n+1]}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A086645, A375450, A375451.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Sep 11 2024