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%I #11 Mar 11 2018 20:05:36
%S 1,1,8,261,18400,2383800,523322208,179218487938,89683747547520,
%T 62708998782383118,59089857758999295000,72802718522742159383487,
%U 114483047893936622981751744,225239782463528942954480067562,545115039824232864946208073815480,1599336782706815943896682999835671375,5617211714811070144606976491531059082752
%N G.f. A(x) satisfies: [x^n] A( x/A(x)^(2*n^2) ) = 0 for n>1.
%C Conjecture: n^2 divides a(n) for n>=1.
%H Paul D. Hanna, <a href="/A300734/b300734.txt">Table of n, a(n) for n = 0..200</a>
%e The table of coefficients in A(x/A(x)^(2*n^2)) begins:
%e n=0: [1, 1, 8, 261, 18400, 2383800, 523322208, 179218487938, ...];
%e n=1: [1, 1, 6, 216, 16180, 2187188, 493439464, 171718406096, ...];
%e n=2: [1, 1, 0, 105, 10568, 1672006, 412368512, 150825821184, ...];
%e n=3: [1, 1, -10, 0, 4228, 1024056, 301838400, 120746446648, ...];
%e n=4: [1, 1, -24, 21, 0, 439248, 186683808, 86739931322, ...];
%e n=5: [1, 1, -42, 336, -1500, 0, 84158448, 53438819528, ...];
%e n=6: [1, 1, -64, 1161, -5240, -403722, 0, 23924633200, ...];
%e n=7: [1, 1, -90, 2760, -26092, -1014644, -61791496, 0, ...];
%e n=8: [1, 1, -120, 5445, -94112, -1888744, -72316832, -16993280158, 0, ...]; ...
%e such that the main diagonal consists of all zeros after the initial terms.
%e The values a(n)/n^2 for n>=1 begin:
%e [1, 2, 29, 1150, 95352, 14536728, 3657520162, 1401308555430, 774185170152878, ...].
%o (PARI) {a(n) = my(A=[1, 1]); for(i=2, n, A=concat(A, 0); A[#A] = -Vec(subst(Ser(A), x, x/Ser(A)^(2*(#A-1)^2)))[#A]); A[n+1]}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A292394.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Mar 11 2018
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