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A295811 G.f. A(x) satisfies: [x^(n-1)] A(x)^(n^2) = 2*n * [x^(n-2)] A(x)^(n^2) for n>=2, with A(0) = 1. 3

%I #20 Mar 14 2018 14:09:45

%S 1,1,2,11,140,2898,80844,2786091,113184008,5266198778,275248731860,

%T 15939117549502,1012084698990904,69901132180300132,

%U 5217426460077854712,418615099531669351443,35942031310982080239120,3289533291926922095871546,319841125714352173292953668,32937612567848507536114539402,3582858531960091228861488651864

%N G.f. A(x) satisfies: [x^(n-1)] A(x)^(n^2) = 2*n * [x^(n-2)] A(x)^(n^2) for n>=2, with A(0) = 1.

%C Compare g.f. to: [x^(n-1)] G(x)^n = 2 * [x^(n-2)] G(x)^n for n>=2 holds when G(x) = 1/(1-x).

%H Paul D. Hanna, <a href="/A295811/b295811.txt">Table of n, a(n) for n = 0..300</a>

%F a(2^k - 1) is odd for k>=0 and a(n) is even elsewhere (conjecture).

%F a(n) ~ c * d^n * n! / n^3, where d = -4/(LambertW(-2*exp(-2))*(2+LambertW(-2*exp(-2)))) = 6.176554609483480358231680164050876553672889794284... and c = 2.719099850893334482... - _Vaclav Kotesovec_, Feb 07 2018

%e G.f.: A(x) = 1 + x + 2*x^2 + 11*x^3 + 140*x^4 + 2898*x^5 + 80844*x^6 + 2786091*x^7 + 113184008*x^8 + 5266198778*x^9 + 275248731860*x^10 + ...

%e ILLUSTRATION OF THE DEFINITION.

%e The table of coefficients of x^k in A(x)^(n^2) begins:

%e n=1: [1, 1, 2, 11, 140, 2898, 80844, ...];

%e n=2: [1, 4, 14, 72, 741, 13724, 364546, ...];

%e n=3: [1, 9, 54, 327, 2826, 42660, 1017720, ...];

%e n=4: [1, 16, 152, 1216, 10540, 129376, 2559792, ...];

%e n=5: [1, 25, 350, 3775, 37750, 427480, 6820800, ...];

%e n=6: [1, 36, 702, 10056, 123165, 1477980, 20712546, ...];

%e n=7: [1, 49, 1274, 23667, 359856, 4953998, 69355972, ...]; ...

%e in which the main diagonal

%e D0 = [1, 4, 54, 1216, 37750, 1477980, 69355972, 3775816704, ...]

%e and the adjacent diagonal

%e D1 = [1, 9, 152, 3775, 123165, 4953998, 235988544, 12954335103, ...]

%e are related by D0[n-1] = 2*n*D1[n-2] for n>=2.

%e The related sequence D0[n-1]/n^2, n>=1, begins:

%e [1, 1, 6, 76, 1510, 41055, 1415428, 58997136, 2878741134, 160698224230, ...].

%o (PARI) {a(n) = my(A=[1]); for(m=2, n+1, A=concat(A, 0); V=Vec(Ser(A)^(m^2)); A[#A] = V[#A-1]*2/m - V[#A]/m^2 ); A[n+1]}

%o for(n=0,20,print1(a(n),", "))

%Y Cf. A295766, A088715, A300873.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Feb 02 2018

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