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A323693 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = (n+1)^2 * [x^(n-1)] A(x)^(n+1) for n >= 1 with A'(0) = 1.

%I A323693 #10 Feb 20 2019 12:38:17
%S A323693 1,2,14,228,6332,255800,13862744,962576816,83146713104,8746885895136,
%T A323693 1102050352603232,163997224386523712,28480503345597714112,
%U A323693 5711832009893579651456,1310680283957123653000064,341305200596595166803458816,100122955976950431349888239872,32871729257928892872345863470592,12007438407819424861612909690881536,4854069613493626427129286480218215424
%N A323693 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = (n+1)^2 * [x^(n-1)] A(x)^(n+1) for n >= 1 with A'(0) = 1.
%C A323693 a(n) / 2^floor((n+1)/2) is odd for n >= 0 (conjecture).
%H A323693 Paul D. Hanna, <a href="/A323693/b323693.txt">Table of n, a(n) for n = 0..300</a>
%e A323693 G.f.: A(x) = 1 + 2*x + 14*x^2 + 228*x^3 + 6332*x^4 + 255800*x^5 + 13862744*x^6 + 962576816*x^7 + 83146713104*x^8 + 8746885895136*x^9 + ...
%e A323693 The table of coefficients of x^k in A(x)^n starts as
%e A323693 n=1: [1, 2, 14, 228, 6332, 255800, 13862744, ...];
%e A323693 n=2: [1, 4, 32, 512, 13772, 543312, 28977968, ...];
%e A323693 n=3: [1, 6, 54, 860, 22488, 866448, 45462704, ...];
%e A323693 n=4: [1, 8, 80, 1280, 32664, 1229568, 63445984, ...];
%e A323693 n=5: [1, 10, 110, 1780, 44500, 1637512, 83069960, ...];
%e A323693 n=6: [1, 12, 144, 2368, 58212, 2095632, 104491088, ...];
%e A323693 n=7: [1, 14, 182, 3052, 74032, 2609824, 127881376, ...]; ...
%e A323693 RELATED SEQUENCES.
%e A323693 In the above table, the main diagonal begins
%e A323693 [1, 4, 54, 1280, 44500, 2095632, 127881376, 9819500544, ...]
%e A323693 which, when divided by (n+1)^2, yields the secondary diagonal (A323694):
%e A323693 [1, 1, 6, 80, 1780, 58212, 2609824, 153429696, 11457990000, ...].
%e A323693 The sequence a(n) / 2^floor((n+1)/2) appears to consist only of odd numbers:
%e A323693 [1, 1, 7, 57, 1583, 31975, 1732843, 60161051, 5196669569, 273340184223, 34439073518851, 2562456631039433, 445007864774964283, ...].
%o A323693 (PARI) {a(n) = my(A=[1], V); for(m=2, n+1, A=concat(A, 0); V=Vec(Ser(A)^m); A[#A] = V[#A-1]*m - V[#A]/m ); A[n+1]}
%o A323693 for(n=0, 20, print1(a(n), ", "))
%o A323693 (PARI) /* Informal method of obtaining N terms: */
%o A323693 N=30; A=[1]; for(n=2, N, A=concat(A, 0); V=Vec(Ser(A)^n); A[#A] = V[#A-1]*n - V[#A]/n ); A
%Y A323693 Cf. A323694, A295766.
%K A323693 nonn
%O A323693 0,2
%A A323693 _Paul D. Hanna_, Feb 20 2019