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.

A360231 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n-1))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 1, 6, 53, 628, 9167, 156309, 3021720, 64960004, 1532234825, 39270176511, 1085601040372, 32185085432757, 1018593646880447, 34279111177431666, 1222648239226278333, 46084480032637208699, 1830881732391546532475, 76488074741796221197580, 3352854778050665597014436
Offset: 0

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Author

Paul D. Hanna, Feb 02 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + x^2 + 6*x^3 + 53*x^4 + 628*x^5 + 9167*x^6 + 156309*x^7 + 3021720*x^8 + 64960004*x^9 + 1532234825*x^10 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 2*x^2 + 10*x^3 + 86*x^4 + 1004*x^5 + 14507*x^6 + 246218*x^7 + 4753205*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(n-1))^(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, 6/3, 40/4, 430/5, 6024/6, 101549/7, 1969744/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  1,   6,   53,  628,   9167,  156309, ...];
n=1: [1, 2,  3,  14,  119, 1374,  19732,  332844, ...];
n=2: [1, 3,  6,  25,  201, 2259,  31891,  531933, ...];
n=3: [1, 4, 10,  40,  303, 3308,  45870,  756192, ...];
n=4: [1, 5, 15,  60,  430, 4551,  61930, 1008565, ...];
n=5: [1, 6, 21,  86,  588, 6024,  80373, 1292370, ...];
n=6: [1, 7, 28, 119,  784, 7770, 101549, 1611352, ...];
n=7: [1, 8, 36, 160, 1026, 9840, 125864, 1969744, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(n-1))^(n+1):
n=0: [1, 1, -1,   0,   -5,   -42,   -528,   -7939, ...];
n=1: [1, 2,  1,   0,    0,     0,      0,       0, ...];
n=2: [1, 3,  6,  10,   30,   207,   2266,   31824, ...];
n=3: [1, 4, 14,  40,  141,   808,   7694,  101288, ...];
n=4: [1, 5, 25, 100,  430,  2376,  19680,  235165, ...];
n=5: [1, 6, 39, 200, 1035,  6024,  45879,  490524, ...];
n=6: [1, 7, 56, 350, 2135, 13601, 101549,  988338, ...];
n=7: [1, 8, 76, 560, 3950, 27888, 213952, 1969744, ...]; ...
to see that the main diagonals of the tables are the same:
[1, 2, 6, 40, 430, 6024, 101549, 1969744, ...].
		

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-1))^(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-1))^(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-1))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alfa, where d = A360279, alfa = 0.311338934287018467072138011497837... and c = 0.1932932528309324180094... - Vaclav Kotesovec, Feb 03 2023