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.

A360235 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n+3))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 5, 48, 673, 12057, 256763, 6232909, 168035350, 4945380012, 157008686993, 5331606427775, 192417007138176, 7344652874314128, 295384546093569838, 12478509340848604628, 552330553975194126634, 25560514938260757190962, 1234444956694450007259989, 62114842767595821207341042
Offset: 0

Views

Author

Paul D. Hanna, Jan 30 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 48*x^3 + 673*x^4 + 12057*x^5 + 256763*x^6 + 6232909*x^7 + 168035350*x^8 + 4945380012*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 6*x^2 + 64*x^3 + 946*x^4 + 17403*x^5 + 375913*x^6 + 9203150*x^7 + 249561291*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(n+3))^(n+1) / (n+1),
as well as b(n) = [x^n] A(x)^(n+1) / (n+1),
so that b(n) begin:
[1/1, 2/2, 18/3, 256/4, 4730/5, 104418/6, 2631391/7, 73625200/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  5,  48,  673,  12057,  256763,  6232909, ...];
n=1: [1, 2, 11, 106, 1467,  25940,  546674, 13164522, ...];
n=2: [1, 3, 18, 175, 2397,  41868,  873317, 20861712, ...];
n=3: [1, 4, 26, 256, 3479,  60080, 1240618, 29397424, ...];
n=4: [1, 5, 35, 350, 4730,  80836, 1652870, 38851165, ...];
n=5: [1, 6, 45, 458, 6168, 104418, 2114759, 49309524, ...];
n=6: [1, 7, 56, 581, 7812, 131131, 2631391, 60866723, ...];
n=7: [1, 8, 68, 720, 9682, 161304, 3208320, 73625200, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(n+3))^(n+1):
n=0: [1, 1,   3,   18,   175,   2397,   41868,   873317, ...];
n=1: [1, 2,   9,   60,   580,   7678,  129842,  2642540, ...];
n=2: [1, 3,  18,  136,  1350,  17520,  287288,  5690016, ...];
n=3: [1, 4,  30,  256,  2661,  34404,  550050, 10593112, ...];
n=4: [1, 5,  45,  430,  4730,  61811,  971600, 18221525, ...];
n=5: [1, 6,  63,  668,  7815, 104418, 1629245, 29869968, ...];
n=6: [1, 7,  84,  980, 12215, 168294, 2631391, 47432554, ...];
n=7: [1, 8, 108, 1376, 18270, 261096, 4125864, 73625200, ...]; ...
to see that the main diagonals of the tables are the same.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(m+3))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n+3))^(n+1) for n >= 0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(n+3))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alfa, where d = A360279 = 2.12460658362428979..., alfa = 4.09132043601400805425594207544980..., c = 0.00160512950354606176706886534963706... - Vaclav Kotesovec, Jan 31 2023