A360581 Expansion of A(x) satisfying [x^n] A(x)^n / (1 + x*A(x)^n)^n = 0 for n > 0.

1, 1, 3, 17, 131, 1204, 12587, 149131, 2036675, 32358153, 587313706, 11761213199, 252859744189, 5785648936988, 141627609404793, 3737907237793369, 106414467836076985, 3241492594168333618, 104522041356412895455, 3541554178675758259947, 125782730912626755808358

Offset: 0

Author

Paul D. Hanna, Mar 12 2023

Keywords

• nonn

Examples

			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 + ...
The table of coefficients in the successive powers of g.f. A(x) begins:
n = 1: [1, 1,  3,  17,  131,  1204,  12587,  149131, ...];
n = 2: [1, 2,  7,  40,  305,  2772,  28657,  335114, ...];
n = 3: [1, 3, 12,  70,  531,  4782,  48936,  565245, ...];
n = 4: [1, 4, 18, 108,  819,  7324,  74272,  848064, ...];
n = 5: [1, 5, 25, 155, 1180, 10501, 105650, 1193530, ...];
n = 6: [1, 6, 33, 212, 1626, 14430, 144208, 1613214, ...];
n = 7: [1, 7, 42, 280, 2170, 19243, 191254, 2120511, ...];
n = 8: [1, 8, 52, 360, 2826, 25088, 248284, 2730872, ...];
...
The table of coefficients in A(x)^n/(1 + x*A(x)^n)^n begins:
n = 1: [1, 0,   2,   12,  100,  955, 10258, 124565, ...];
n = 2: [1, 0,   2,   18,  161, 1606, 17757, 220834, ...];
n = 3: [1, 0,   0,   15,  168, 1806, 21000, 272856, ...];
n = 4: [1, 0,  -4,    0,  114, 1504, 19220, 270692, ...];
n = 5: [1, 0, -10,  -30,    0,  800, 12970, 215445, ...];
n = 6: [1, 0, -18,  -78, -165,    0,  4797, 123990, ...];
n = 7: [1, 0, -28, -147, -364, -329,     0,  32767, ...];
n = 8: [1, 0, -40, -240, -572,  696,  7472,      0, ...];
...
in which the diagonal of all zeros illustrates that
[x^n] A(x)^n / (1 + x*A(x)^n)^n = 0 for n > 0.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A360582, A360583, A360584, A303063.

Programs

PARI

{a(n) = my(A=[1]); for(i=1,n, A = concat(A,0);
A[#A] = -polcoeff( Ser(A)^(#A)/(1 + x*Ser(A)^(#A))^(#A), #A-1)/(#A) );A[n+1]}
for(n=0,30,print1(a(n),", "))

Formula

From Vaclav Kotesovec, Mar 13 2023: (Start)

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

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