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.

A113867 a(n) = a(n-1) + 2^(A047258(n)) for n>1, a(1)=1.

Original entry on oeis.org

1, 17, 49, 113, 1137, 3185, 7281, 72817, 203889, 466033, 4660337, 13048945, 29826161, 298261617, 835132529, 1908874353, 19088743537, 53448481905, 122167958641, 1221679586417, 3420702841969, 7818749353073, 78187493530737
Offset: 1

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Author

Artur Jasinski, Jan 27 2006

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 + 16 x + 32 x^2) / ((-1 + x) (- 1 + 4 x) (1 + 4 x + 16 x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, May 20 2013 *)

Formula

G.f.: x*(1+16*x+32*x^2)/((-1+x)*(-1+4*x)*(1+4*x+16*x^2)). - Vaclav Kotesovec, Nov 28 2012

Extensions

Edited with better definition and offset corrected by Omar E. Pol, Jan 08 2009

A047245 Numbers that are congruent to {1, 2, 3} mod 6.

Original entry on oeis.org

1, 2, 3, 7, 8, 9, 13, 14, 15, 19, 20, 21, 25, 26, 27, 31, 32, 33, 37, 38, 39, 43, 44, 45, 49, 50, 51, 55, 56, 57, 61, 62, 63, 67, 68, 69, 73, 74, 75, 79, 80, 81, 85, 86, 87, 91, 92, 93, 97, 98, 99, 103, 104, 105, 109, 110, 111, 115, 116, 117, 121, 122, 123
Offset: 1

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Author

Keywords

Comments

a(k)^m is a term iff {a(k) is odd and m is a nonnegative integer} or {m is in A004273}. - Jerzy R Borysowicz, May 08 2023

Crossrefs

Cf. A047240, A047244, A047258 (complement).

Programs

Formula

From Johannes W. Meijer, Jun 07 2011: (Start)
a(n) = ceiling(n/3) + ceiling((n-1)/3) + ceiling((n-2)/3) + 3*ceiling((n-3)/3).
G.f.: x*(1+x+x^2+3*x^3)/((x-1)^2*(x^2+x+1)). (End)
a(n) = 3*floor((n-1)/3) + n. - Gary Detlefs, Dec 22 2011
From Wesley Ivan Hurt, Apr 13 2015: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = 2*n-3 + ((2*n-3) mod 3). (End)
From Wesley Ivan Hurt, Jun 13 2016: (Start)
a(n) = 2*n - 2 - cos(2*n*Pi/3) + sin(2*n*Pi/3)/sqrt(3).
a(3k) = 6k-3, a(3k-1) = 6k-4, a(3k-2) = 6k-5. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = (9-2*sqrt(3))*Pi/36 + log(2+sqrt(3))/(2*sqrt(3)) - log(2)/6. - Amiram Eldar, Dec 14 2021
Showing 1-2 of 2 results.