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.

Showing 1-2 of 2 results.

A305492 a(n) = ((1 + y)^n - (1 - y)^n)/y with y = sqrt(8).

Original entry on oeis.org

0, 2, 4, 22, 72, 298, 1100, 4286, 16272, 62546, 238996, 915814, 3504600, 13419898, 51371996, 196683278, 752970528, 2882724002, 11036241700, 42251551414, 161756794728, 619274449354, 2370846461804, 9076614069086, 34749153370800
Offset: 0

Views

Author

Peter Luschny, Jun 02 2018

Keywords

Examples

			Array ((1+y)^n - (1-y)^n)/y with y = sqrt(k).
[k\n]
[1]   1, 2, 4,  8, 16, 32,   64,  128,    256,   512,   1024, ...
[2]   0, 2, 4, 10, 24, 58,  140,  338,    816,  1970,   4756, ...
[3]   0, 2, 4, 12, 32, 88,  240,  656,   1792,  4896,  13376, ...
[4]   0, 2, 4, 14, 40, 122, 364,  1094,  3280,  9842,  29524, ...
[5]   0, 2, 4, 16, 48, 160, 512,  1664,  5376, 17408,  56320, ...
[6]   0, 2, 4, 18, 56, 202, 684,  2378,  8176, 28242,  97364, ...
[7]   0, 2, 4, 20, 64, 248, 880,  3248, 11776, 43040, 156736, ...
[8]   0, 2, 4, 22, 72, 298, 1100, 4286, 16272, 62546, 238996, ...
[9]   0, 2, 4, 24, 80, 352, 1344, 5504, 21760, 87552, 349184, ...
		

Crossrefs

Let f(n, y) = ((1 + y)^n - (1 - y)^n)/y.
f(n, 1 ) = A000079(n);
f(n, sqrt(2)) = A163271(n+1);
f(n, sqrt(3)) = A028860(n+2);
f(n, 2 ) = A152011(n) for n>0;
f(n, sqrt(5)) = A103435(n);
f(n, sqrt(6)) = A083694(n);
f(n, sqrt(7)) = A274520(n);
f(n, sqrt(8)) = a(n);
f(n, 3 ) = A192382(n+1);
Cf. A305491.
Equals 2 * A015519.

Programs

  • Maple
    egf :=  (n,x) -> 2*exp(x)*sinh(sqrt(n)*x)/sqrt(n):
    ser := series(egf(8,x), x, 26):
    seq(n!*coeff(ser,x, n), n=0..24);
  • Mathematica
    Table[Simplify[((1 + Sqrt[8])^n - (1 - Sqrt[8])^n)/ Sqrt[8]], {n, 0, 24}]
  • PARI
    concat(0, Vec(2*x / (1 - 2*x - 7*x^2) + O(x^40))) \\ Colin Barker, Jun 05 2018

Formula

E.g.f.: 2*exp(x)*sinh(sqrt(n)*x)/sqrt(n) for n = 8.
From Colin Barker, Jun 02 2018: (Start)
G.f.: 2*x / (1 - 2*x - 7*x^2).
a(n) = 2*a(n-1) + 7*a(n-2) for n>1.
(End)

A277091 a(n) = ((1 + sqrt(15))^n - (1 - sqrt(15))^n)/sqrt(15).

Original entry on oeis.org

0, 2, 4, 36, 128, 760, 3312, 17264, 80896, 403488, 1939520, 9527872, 46209024, 225808256, 1098542848, 5358401280, 26096402432, 127210422784, 619770479616, 3020486878208, 14717760471040, 71722337236992, 349493321068544, 1703099363454976, 8299105221869568, 40441601532108800
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 29 2016

Keywords

Comments

Number of zeros in substitution system {0 -> 1111111, 1 -> 1001} at step n from initial string "1" (see example).

Examples

			Evolution from initial string "1": 1 -> 1001 -> 1001111111111111111001 -> ...
Therefore, number of zeros at step n:
a(0) = 0;
a(1) = 2;
a(2) = 4, etc.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{2, 14}, {0, 2}, 26]
  • PARI
    concat(0, Vec(2*x/(1-2*x-14*x^2) + O(x^99))) \\ Altug Alkan, Oct 01 2016

Formula

O.g.f.: 2*x/(1 - 2*x - 14*x^2).
E.g.f.: 2*sinh(sqrt(15)*x)*exp(x)/sqrt(15).
a(n) = 2*a(n-1) + 14*a(n-2).
Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(15) = 1 + A010472.
Showing 1-2 of 2 results.