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Expansion of A(x) satisfying [x^n] A(x)^n / (1 + x*A(x)^n)^n = 0 for n > 0.
3

%I #9 Sep 13 2023 10:49:00

%S 1,1,3,17,131,1204,12587,149131,2036675,32358153,587313706,

%T 11761213199,252859744189,5785648936988,141627609404793,

%U 3737907237793369,106414467836076985,3241492594168333618,104522041356412895455,3541554178675758259947,125782730912626755808358

%N Expansion of A(x) satisfying [x^n] A(x)^n / (1 + x*A(x)^n)^n = 0 for n > 0.

%H Paul D. Hanna, <a href="/A360581/b360581.txt">Table of n, a(n) for n = 0..300</a>

%F From _Vaclav Kotesovec_, Mar 13 2023: (Start)

%F a(n) ~ c * n! * n^alpha / LambertW(1)^n, where alpha = 0.33953... and c = 0.1881608377753...

%F Conjecture: alpha = 3*LambertW(1) - 2 + 1/(1 + LambertW(1)) = 0.33953361459446... (End)

%e G.f. A(x) = 1 + x + 3*x^2 + 17*x^3 + 131*x^4 + 1204*x^5 + 12587*x^6 + 149131*x^7 + 2036675*x^8 + 32358153*x^9 + 587313706*x^10 + ...

%e The table of coefficients in the successive powers of g.f. A(x) begins:

%e n = 1: [1, 1, 3, 17, 131, 1204, 12587, 149131, ...];

%e n = 2: [1, 2, 7, 40, 305, 2772, 28657, 335114, ...];

%e n = 3: [1, 3, 12, 70, 531, 4782, 48936, 565245, ...];

%e n = 4: [1, 4, 18, 108, 819, 7324, 74272, 848064, ...];

%e n = 5: [1, 5, 25, 155, 1180, 10501, 105650, 1193530, ...];

%e n = 6: [1, 6, 33, 212, 1626, 14430, 144208, 1613214, ...];

%e n = 7: [1, 7, 42, 280, 2170, 19243, 191254, 2120511, ...];

%e n = 8: [1, 8, 52, 360, 2826, 25088, 248284, 2730872, ...];

%e ...

%e The table of coefficients in A(x)^n/(1 + x*A(x)^n)^n begins:

%e n = 1: [1, 0, 2, 12, 100, 955, 10258, 124565, ...];

%e n = 2: [1, 0, 2, 18, 161, 1606, 17757, 220834, ...];

%e n = 3: [1, 0, 0, 15, 168, 1806, 21000, 272856, ...];

%e n = 4: [1, 0, -4, 0, 114, 1504, 19220, 270692, ...];

%e n = 5: [1, 0, -10, -30, 0, 800, 12970, 215445, ...];

%e n = 6: [1, 0, -18, -78, -165, 0, 4797, 123990, ...];

%e n = 7: [1, 0, -28, -147, -364, -329, 0, 32767, ...];

%e n = 8: [1, 0, -40, -240, -572, 696, 7472, 0, ...];

%e ...

%e in which the diagonal of all zeros illustrates that

%e [x^n] A(x)^n / (1 + x*A(x)^n)^n = 0 for n > 0.

%o (PARI) {a(n) = my(A=[1]); for(i=1,n, A = concat(A,0);

%o A[#A] = -polcoeff( Ser(A)^(#A)/(1 + x*Ser(A)^(#A))^(#A), #A-1)/(#A) );A[n+1]}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A360582, A360583, A360584, A303063.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Mar 12 2023