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%I #22 Oct 20 2020 03:26:57
%S 1,1,2,19,648,35034,2670208,272185631,35771059816,5898174461866,
%T 1193978013414012,291425454605906442,84460020885960997128,
%U 28684514514013845794992,11286674783772300965795444,5094396145960067896123076579,2614825930760477198096727703504,1514515134552667357882578305228354
%N G.f. A(x) satisfies: [x^n] A(x)^(n*(n+1)) / (x*A(x)^(n+1))' = 0 for n>1.
%C Odd terms appear to occur only at positions 2^n - 1 for n>=0.
%C Compare to: [x^n] (x*F(x)^n)' / F(x)^(n*(n+1)) = 0 for n>1 holds when F(0) = 1.
%H Paul D. Hanna, <a href="/A300995/b300995.txt">Table of n, a(n) for n = 0..300</a>
%F G.f. A(x) satisfies: [x^n] A(x)^(n^2) / (A(x) + (n+1)*x*A'(x)) = 0 for n>1.
%F a(n) ~ c * 2^n * n!^2, where c = 0.093409272388974721425474653920126051... - _Vaclav Kotesovec_, Oct 20 2020
%e G.f.: A(x) = 1 + x + 2*x^2 + 19*x^3 + 648*x^4 + 35034*x^5 + 2670208*x^6 + 272185631*x^7 + 35771059816*x^8 + 5898174461866*x^9 + ...
%e such that [x^n] A(x)^(n*(n+1)) / (x*A(x)^(n+1))' = 0 for n>1.
%e ILLUSTRATION OF DEFINITION.
%e The table of coefficients in A(x)^(n*(n+1)) / (x*A(x)^(n+1))' begins:
%e n=0: [1, -2, -2, -60, -2956, -197300, -17847672, -2102383528, ...];
%e n=1: [1, -2, -2, -88, -4634, -323628, -30231156, -3645343392, ...];
%e n=2: [1, 0, 0, -86, -5211, -389844, -37957504, -4706848080, ...];
%e n=3: [1, 4, 16, 0, -4318, -390512, -40986336, -5301293056, ...];
%e n=4: [1, 10, 70, 400, 0, -294292, -38253360, -5373817820, ...];
%e n=5: [1, 18, 198, 1760, 15084, 0, -26932912, -4775237568, ...];
%e n=6: [1, 28, 448, 5502, 63581, 818104, 0, -3197458336, ...];
%e n=7: [1, 40, 880, 14304, 204524, 3166480, 61757056, 0, ...];
%e n=8: [1, 54, 1566, 32700, 572292, 9885564, 214256808, 6302260080, 0, ...]; ...
%e in which the main diagonal consists of all zeros after the initial terms, illustrating that [x^n] A(x)^(n*(n+1)) / (x*A(x)^(n+1))' = 0 for n>1.
%e RELATED SERIES.
%e log(A(x)) = x + 3*x^2/2 + 52*x^3/3 + 2515*x^4/4 + 171846*x^5/5 + 15806406*x^6/6 + 1886292528*x^7/7 + 283963660371*x^8/8 + 52758607655410*x^9/9 + ...
%o (PARI) {a(n) = my(A=[1, 1]); for(i=1, n, A = concat(A, 0); A[#A] = Vec( Ser(A)^(#A*(#A-1))/(x*Ser(A)^(#A))' )[#A]/(#A)); A[n+1]}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A302060, A300994, A300627, A302059.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Mar 29 2018
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