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 1-6 of 6 results.

A167762 a(n) = 2*a(n-1)+3*a(n-2)-6*a(n-3) starting a(0)=a(1)=0, a(2)=1.

Original entry on oeis.org

0, 0, 1, 2, 7, 14, 37, 74, 175, 350, 781, 1562, 3367, 6734, 14197, 28394, 58975, 117950, 242461, 484922, 989527, 1979054, 4017157, 8034314, 16245775, 32491550, 65514541, 131029082, 263652487, 527304974, 1059392917, 2118785834, 4251920575, 8503841150
Offset: 0

Views

Author

Paul Curtz, Nov 11 2009

Keywords

Comments

Inverse binomial transform yields two zeros followed by A077917 (a signed variant of A127864).
a(n) mod 10 is zero followed by a sequence with period length 8: 0, 1, 2, 7, 4, 7, 4, 5 (repeat).
a(n) is the number of length n+1 binary words with some prefix w such that w contains three more 1's than 0's and no prefix of w contains three more 0's than 1's. - Geoffrey Critzer, Dec 13 2013
From Gus Wiseman, Oct 06 2023: (Start)
Also the number of subsets of {1..n} with two distinct elements summing to n + 1. For example, the a(2) = 1 through a(5) = 14 subsets are:
{1,2} {1,3} {1,4} {1,5}
{1,2,3} {2,3} {2,4}
{1,2,3} {1,2,4}
{1,2,4} {1,2,5}
{1,3,4} {1,3,5}
{2,3,4} {1,4,5}
{1,2,3,4} {2,3,4}
{2,4,5}
{1,2,3,4}
{1,2,3,5}
{1,2,4,5}
{1,3,4,5}
{2,3,4,5}
{1,2,3,4,5}
The complement is counted by A038754.
Allowing twins gives A167936, complement A108411.
For n instead of n + 1 we have A365544, complement A068911.
The version for all subsets (not just pairs) is A366130.
(End)

Crossrefs

First differences are A167936, complement A108411.

Programs

  • Mathematica
    LinearRecurrence[{2,3,-6},{0,0,1},40] (* Harvey P. Dale, Sep 17 2013 *)
    CoefficientList[Series[x^2/((2 x - 1) (3 x^2 - 1)), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 17 2013 *)
    Table[Length[Select[Subsets[Range[n]],MemberQ[Total/@Subsets[#,{2}],n+1]&]],{n,0,10}] (* Gus Wiseman, Oct 06 2023 *)

Formula

a(n) mod 9 = A153130(n), n>3 (essentially the same as A154529, A146501 and A029898).
a(n+1)-2*a(n) = 0 if n even, = A000244((1+n)/2) if n odd.
a(2*n) = A005061(n). a(2*n+1) = 2*A005061(n).
G.f.: x^2/((2*x-1)*(3*x^2-1)). a(n) = 2^n - A038754(n). - R. J. Mathar, Nov 12 2009
G.f.: x^2/(1-2*x-3*x^2+6*x^3). - Philippe Deléham, Nov 11 2009

Extensions

Edited and extended by R. J. Mathar, Nov 12 2009

A154127 Period 6: repeat [1, 2, 5, 8, 7, 4].

Original entry on oeis.org

1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2, 5, 8, 7, 4, 1, 2
Offset: 0

Views

Author

Paul Curtz, Jan 05 2009

Keywords

Crossrefs

Programs

Formula

From R. J. Mathar, Feb 25 2009, Mar 09 2009: (Start)
a(n) = a(n-1) - a(n-3) + a(n-4) for n>3.
G.f.: (1+x+3*x^2+4*x^3)/((1-x)*(1+x)*(x^2-x+1)). (End)
a(n) = (27-cos(n*Pi)-20*cos(n*Pi/3)-4*sqrt(3)*sin(n*Pi/3))/6. - Wesley Ivan Hurt, Jun 17 2016

Extensions

Corrected numerator in g.f R. J. Mathar, Mar 09 2009

A154870 Period 6: repeat [7, 5, 1, -7, -5, -1].

Original entry on oeis.org

7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1, -7, -5, -1, 7, 5, 1
Offset: 0

Views

Author

Paul Curtz, Jan 16 2009

Keywords

Comments

The sequence b(n) = (-A153130(n)) mod 9 = A153130(n+3) = A146501(n-1) = 8, 7, 5, 1, 2, 4,... has period length 6. This here is a(n)=b(n)-A153130(n).
a(n) is (-1)^(n+1) * numerator of F(n) where F(n) = f(F(n-1)) starting from F(0) = -7/4 and step f(z) = z^2 -29/16. - Nicolas Bělohoubek, Nov 20 2024

Crossrefs

Programs

Formula

a(n) = -a(n-3) for n>2; G.f.: (7+5*x+x^2)/((1+x)*(1-x+x^2)). [R. J. Mathar, Jan 23 2009]
a(n) = cos(n*Pi) + 6*cos(n*Pi/3) + 2*sqrt(3)*sin(n*Pi/3). - Wesley Ivan Hurt, Jun 20 2016

Extensions

Edited and extended by R. J. Mathar, Jan 23 2009

A167784 a(n) = 2^n - (1 - (-1)^n)*3^((n-1)/2).

Original entry on oeis.org

1, 0, 4, 2, 16, 14, 64, 74, 256, 350, 1024, 1562, 4096, 6734, 16384, 28394, 65536, 117950, 262144, 484922, 1048576, 1979054, 4194304, 8034314, 16777216, 32491550, 67108864, 131029082, 268435456, 527304974, 1073741824, 2118785834, 4294967296, 8503841150
Offset: 0

Views

Author

Paul Curtz, Nov 12 2009

Keywords

Comments

Binomial transform of A077917, the signed variant of A127864.

Crossrefs

Cf. A154383.

Programs

  • Maple
    seq(2^n - (1 - (-1)^n)*3^((n-1)/2), n=0..100); # Robert Israel, Apr 11 2019
  • Mathematica
    LinearRecurrence[{2, 3, -6}, {1, 0, 4}, 40] (* Harvey P. Dale, Nov 29 2011 *)

Formula

a(n) = A167936(n+1) - A167936(n).
a(2n) = A000302(n). a(2n+1) = 2*A005061(n).
a(n) = 2*a(n-1) + 3*a(n-2) - 6*a(n-3).
G.f.: (x-1)^2/((2*x-1)*(3*x^2-1)).
a(n+4) mod 9 = A153130(n+4) = A146501(n+2), n>=0.
E.g.f.: exp(2*x) - (2/sqrt(3))*sinh(sqrt(3)*x). - G. C. Greubel, Jun 27 2016

Extensions

Edited and extended by R. J. Mathar, Feb 27 2010
Incorrect b-file corrected by Robert Israel, Apr 11 2019

A166577 Inverse binomial transform of A166517.

Original entry on oeis.org

1, 4, -5, 10, -20, 40, -80, 160, -320, 640, -1280, 2560, -5120, 10240, -20480, 40960, -81920, 163840, -327680, 655360, -1310720, 2621440, -5242880, 10485760, -20971520, 41943040, -83886080, 167772160, -335544320, 671088640, -1342177280, 2684354560, -5368709120
Offset: 0

Views

Author

Paul Curtz, Oct 17 2009

Keywords

Comments

The definition assumes that the offset of A166517 is changed to 0.
A166517 mod 9 yields a periodic sequence with period 1, 5, 4, 8, 7, 2.
This set of numbers in the period appears also in A153130, A146501, and A166304.

Crossrefs

Programs

  • Mathematica
    Join[{1,4},NestList[-2#&,-5,40]] (* Harvey P. Dale, Aug 02 2012 *)
    Join[{1, 4}, LinearRecurrence[{-2}, {-5}, 48]] (* G. C. Greubel, May 17 2016 *)

Formula

a(n) = -2*a(n-1), n>2.
a(n) = (-1)^(n+1)*A020714(n-2), n>1.
From Colin Barker, Jan 07 2013: (Start)
a(n) = -5*(-1)^n*2^(n-2) for n>1.
G.f.: (3*x^2+6*x+1)/(2*x+1). (End)
E.g.f.: (9/4) + (3/2)*x - (5/4)*exp(-2*x). - Alejandro J. Becerra Jr., Feb 15 2021

Extensions

Edited, comments not concerning this sequence removed, and extended by R. J. Mathar, Oct 21 2009

A154529 A090040 mod 9.

Original entry on oeis.org

1, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8
Offset: 0

Views

Author

Paul Curtz, Jan 11 2009

Keywords

Comments

For n>2, equal to 2^(n-2) mod 9 [From Michael B. Porter, Feb 02 2010]
Apart from leading terms the same as A146501, A153130 and A029898. [From R. J. Mathar, Apr 13 2010]

Programs

  • Mathematica
    Join[{1},LinearRecurrence[{1,0,-1,1},{5,1,2,4},101]] (* Ray Chandler, Jul 15 2015 *)

Formula

a(n)=a(n-1)-a(n-3)+a(n-4), n>4. G.f.: (6*x^4+2*x^3+4*x+1-4*x^2)/((1-x)*(1+x)*(x^2-x+1)). [From R. J. Mathar, Feb 25 2009]

Extensions

Edited by N. J. A. Sloane, Jan 12 2009
Extended by Ray Chandler, Jul 15 2015
Showing 1-6 of 6 results.