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.

A360347 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n+3))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 7, 105, 2366, 68776, 2390230, 95166058, 4228436480, 206090296497, 10887958126763, 618187895371965, 37479711430699245, 2414492049517400164, 164626564026042206780, 11841796830661101527618, 896184803460067359995232, 71189783172592806474895908
Offset: 0

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Author

Paul D. Hanna, Feb 05 2023

Keywords

Comments

Sequences with g.f. A(x,k) such that [x^n] A(x,k)^(n+1) = [x^n] (1 + x*A(x,k)^(2*n+k))^(n+1) have a rate of growth: a(n) ~ c(k) * d^n * n! * n^alpha(k), where d = 3.93464558322824528799... (independent on k) and alpha(k) = 1.1169011279372... + k*0.518500901361... - Vaclav Kotesovec, Feb 06 2023

Examples

			G.f.: A(x) = 1 + x + 7*x^2 + 105*x^3 + 2366*x^4 + 68776*x^5 + 2390230*x^6 + 95166058*x^7 + 4228436480*x^8 + 206090296497*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 8*x^2 + 127*x^3 + 2927*x^4 + 85892*x^5 + 2998264*x^6 + 119665415*x^7 + 5325877575*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(2*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, 24/3, 508/4, 14635/5, 515352/6, 20987848/7, 957323320/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  7,  105,  2366,  68776,  2390230,  95166058, ...];
n=1: [1, 2, 15,  224,  4991, 143754,  4962161, 196572300, ...];
n=2: [1, 3, 24,  358,  7896, 225396,  7727644, 304572936, ...];
n=3: [1, 4, 34,  508, 11103, 314192, 10699244, 419541832, ...];
n=4: [1, 5, 45,  675, 14635, 410661, 13890275, 541873525, ...];
n=5: [1, 6, 57,  860, 18516, 515352, 17314836, 671984280, ...];
n=6: [1, 7, 70, 1064, 22771, 628845, 20987848, 810313190, ...];
n=7: [1, 8, 84, 1288, 27426, 751752, 24925092, 957323320, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(2*n+3))^(n+1):
n=0: [1, 1,   3,   24,   358,   7896,   225396,   7727644, ...];
n=1: [1, 2,  11,  100,  1465,  31070,   859367,  28808972, ...];
n=2: [1, 3,  24,  253,  3780,  77994,  2089024,  68277867, ...];
n=3: [1, 4,  42,  508,  7915, 161316,  4196916, 133476480, ...];
n=4: [1, 5,  65,  890, 14635, 298981,  7602705, 235213110, ...];
n=5: [1, 6,  93, 1424, 24858, 515352, 12914214, 389369448, ...];
n=6: [1, 7, 126, 2135, 39655, 842331, 20987848, 619044602, ...];
n=7: [1, 8, 164, 3048, 60250, 1320480, 32998388, 957323320, ...]; ...
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)^(2*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)^(2*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)^(2*n+3))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 2.672403832022..., c = 0.0085435225111... - Vaclav Kotesovec, Feb 06 2023