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%I #15 May 29 2024 11:24:33
%S 1,1,13,1333,438073,328561681,482408372341,1262989939509733,
%T 5507311107090685873,37883505322347710775553,
%U 393149949374099099160049501,5930998808712507352448964186421,126060064477829234977371818938653673,3675839897921109642941288187056728970833,143727814785299582494066294788162327508528453
%N E.g.f. A(x) satisfies: [x^n] A(x)^(n^2) = n^3 * [x^(n-1)] A(x)^(n^2) for n>=1.
%C Compare e.g.f. to: [x^n] exp(x)^(n^2) = n * [x^(n-1)] exp(x)^(n^2) for n>=1.
%H Paul D. Hanna, <a href="/A300592/b300592.txt">Table of n, a(n) for n = 0..200</a>
%F E.g.f. A(x) satisfies: log(A(x)) = Sum_{n>=1} A300593(n)*x^n, a power series in x with integer coefficients.
%F a(n) ~ c * n!^4, where c = 3.1056678107899395562612789210816... - _Vaclav Kotesovec_, Oct 14 2020
%e E.g.f.: A(x) = 1 + x + 13*x^2/2! + 1333*x^3/3! + 438073*x^4/4! + 328561681*x^5/5! + 482408372341*x^6/6! + 1262989939509733*x^7/7! + 5507311107090685873*x^8/8! + 37883505322347710775553*x^9/9! + ...
%e ILLUSTRATION OF DEFINITION.
%e The table of coefficients of x^k in A(x)^(n^2) begins:
%e n=1: [(1), (1), 13/2, 1333/6, 438073/24, 328561681/120, ...];
%e n=2: [1, (4), (32), 2912/3, 228032/3, 167874308/15, ...];
%e n=3: [1, 9, (189/2), (5103/2), 1468467/8, 1045214163/40, ...];
%e n=4: [1, 16, 224, (17024/3), (1089536/3), 735471632/15, ...];
%e n=5: [1, 25, 925/2, 70525/6, (15835225/24), (1979403125/24), ...];
%e n=6: [1, 36, 864, 23328, 1161792, (654796044/5), (141435945504/5), ...]; ...
%e in which the coefficients in parenthesis are related by
%e 1 = 1*1; 32 = 2^3*4; 5103/2 = 3^3*189/2; 1089536/3 = 4^3*17024/3; ...
%e illustrating that: [x^n] A(x)^(n^2) = n^3 * [x^(n-1)] A(x)^(n^2).
%e LOGARITHMIC PROPERTY.
%e The logarithm of the e.g.f. is the integer series:
%e log(A(x)) = x + 6*x^2 + 216*x^3 + 18016*x^4 + 2718575*x^5 + 667151244*x^6 + 249904389518*x^7 + 136335045655680*x^8 + 104258627494173747*x^9 + 108236370325030253850*x^10 + 148475074256982964816314*x^11 + 263023328027145941803648512*x^12 + ... + A300593(n)*x^n + ...
%o (PARI) {a(n) = my(A=[1]); for(i=1, n+1, A=concat(A, 0); V=Vec(Ser(A)^((#A-1)^2)); A[#A] = ((#A-1)^3*V[#A-1] - V[#A])/(#A-1)^2 ); n!*A[n+1]}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A182962, A296170, A300590, A300593, A300594.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Mar 09 2018