cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A360583 Expansion of A(x) satisfying [x^n] A(x) / (1 + x*A(x)^(n+1)) = 0 for n > 0.

Original entry on oeis.org

1, 1, 3, 17, 139, 1455, 18326, 267700, 4426686, 81455357, 1646941293, 36238989035, 861298646217, 21978627323651, 599195351716464, 17379759869328515, 534392854606942358, 17363705283005593096, 594513962872698955686, 21395116871674867310280, 807416275879430472577570
Offset: 0

Views

Author

Paul D. Hanna, Mar 12 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 17*x^3 + 139*x^4 + 1455*x^5 + 18326*x^6 + 267700*x^7 + 4426686*x^8 + 81455357*x^9 + 1646941293*x^10 + ...
The table of coefficients in the successive powers of g.f. A(x) begins:
n = 1: [1, 1,  3,  17,  139,  1455,  18326, ...];
n = 2: [1, 2,  7,  40,  321,  3290,  40685, ...];
n = 3: [1, 3, 12,  70,  555,  5583,  67827, ...];
n = 4: [1, 4, 18, 108,  851,  8424, 100624, ...];
n = 5: [1, 5, 25, 155, 1220, 11916, 140085, ...];
n = 6: [1, 6, 33, 212, 1674, 16176, 187372, ...];
n = 7: [1, 7, 42, 280, 2226, 21336, 243817, ...];
...
The table of coefficients in A(x)/(1 + x*A(x)^(n+1)) begins:
n = 1: [1, 0,  1,   9,   88, 1021, 13736, 209940, ...];
n = 2: [1, 0,  0,   5,   64,  821, 11670, 184622, ...];
n = 3: [1, 0, -1,   0,   35,  587,  9283, 155666, ...];
n = 4: [1, 0, -2,  -6,    0,  315,  6555, 122855, ...];
n = 5: [1, 0, -3, -13,  -42,    0,  3467,  86025, ...];
n = 6: [1, 0, -4, -21,  -92, -364,     0,  45079, ...];
n = 7: [1, 0, -5, -30, -151, -784, -3866,      0, ...];
...
in which the diagonal of all zeros illustrates that
[x^n] A(x) / (1 + x*A(x)^(n+1)) = 0 for n > 0.
		

Crossrefs

Programs

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

Formula

a(n) ~ c * n! * n^(3*LambertW(1) - 1 + 1/(1 + LambertW(1))) / LambertW(1)^n, where c = 0.0685246538334926708088... - Vaclav Kotesovec, Mar 13 2023