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.

A360346 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(2*n+2))^(n+1) for n >= 0.

Original entry on oeis.org

1, 1, 6, 82, 1724, 47223, 1555047, 58892186, 2496826094, 116434989450, 5900151126856, 322048641354617, 18810964989814291, 1169843128503194025, 77145176721564799777, 5376524285402806746719, 394887654026596322701724, 30489608056346314234108286, 2469347798211941105406473481
Offset: 0

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Author

Paul D. Hanna, Feb 05 2023

Keywords

Examples

			G.f.: A(x) = 1 + x + 6*x^2 + 82*x^3 + 1724*x^4 + 47223*x^5 + 1555047*x^6 + 58892186*x^7 + 2496826094*x^8 + 116434989450*x^9 + ...
RELATED SERIES.
G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins:
B(x) = 1 + x + 7*x^2 + 101*x^3 + 2161*x^4 + 59544*x^5 + 1965132*x^6 + 74504861*x^7 + 3161424763*x^8 + ... + b(n)*x^n + ...
such that b(n) = [x^n] (1 + x*A(x)^(2*n+2))^(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, 21/3, 404/4, 10805/5, 357264/6, 13755924/7, 596038888/8, ...].
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
n=0: [1, 1,  6,   82,  1724,  47223,  1555047,  58892186, ...];
n=1: [1, 2, 13,  176,  3648,  98878,  3231952, 121743878, ...];
n=2: [1, 3, 21,  283,  5790, 155319,  5039055, 188787837, ...];
n=3: [1, 4, 30,  404,  8169, 216924,  6985240, 260270488, ...];
n=4: [1, 5, 40,  540, 10805, 284096,  9079965, 336452690, ...];
n=5: [1, 6, 51,  692, 13719, 357264, 11333293, 417610542, ...];
n=6: [1, 7, 63,  861, 16933, 436884, 13755924, 504036226, ...];
n=7: [1, 8, 76, 1048, 20470, 523440, 16359228, 596038888, ...]; ...
Compare to the table of coefficients in (1 + x*A(x)^(2*n+2))^(n+1):
n=0: [1, 1,   2,   13,   176,   3648, 98878, 3231952, ...];
n=1: [1, 2,   9,   68,   884,  17386,   454318,  14493920, ...];
n=2: [1, 3,  21,  190,  2508,  47406,  1190949,  36928479, ...];
n=3: [1, 4,  38,  404,  5585, 103464,  2504568,  75227160, ...];
n=4: [1, 5,  60,  735, 10805, 200001,  4698210, 136509465, ...];
n=5: [1, 6,  87, 1208, 19011, 357264,  8227591, 231595008, ...];
n=6: [1, 7, 119, 1848, 31199, 602427, 13755924, 376756199, ...];
n=7: [1, 8, 156, 2680, 48518, 970712, 22218108, 596038888, ...]; ...
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+2))^(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+2))^(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+2))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alpha, where d = 3.93464558322824528799..., alpha = 2.153902930660..., c = 0.01676305987174... - Vaclav Kotesovec, Feb 06 2023