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.

Previous Showing 21-26 of 26 results.

A015609 a(n) = 11*a(n-1) + 12*a(n-2).

Original entry on oeis.org

0, 1, 11, 133, 1595, 19141, 229691, 2756293, 33075515, 396906181, 4762874171, 57154490053, 685853880635, 8230246567621, 98762958811451, 1185155505737413, 14221866068848955, 170662392826187461
Offset: 0

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Author

Keywords

Comments

Number of walks of length n between any two distinct nodes of the complete graph K_13. Example: a(2)=11 because the walks of length 2 between the nodes A and B of the complete graph ABCDEFGHIJKLM are ACB, ADB, AEB, AFB, AGB, AHB, AIB, AJB, AKB, ALB and AMB. - Emeric Deutsch, Apr 01 2004

Crossrefs

Programs

  • Magma
    [(1/13)*(12^n-(-1)^n): n in [0..20]]; // Vincenzo Librandi, Oct 11 2011
    
  • Mathematica
    CoefficientList[Series[x/(1-11*x-12*x^2), {x, 0, 50}], x] (* or *) LinearRecurrence[{11,12}, {0,1}, 30] (* G. C. Greubel, Dec 30 2017 *)
  • PARI
    x='x+O('x^30); concat([0], Vec(x/(1-11*x-12*x^2))) \\ G. C. Greubel, Dec 30 2017
  • Sage
    [lucas_number1(n,11,-12) for n in range(0, 18)] # Zerinvary Lajos, Apr 27 2009
    
  • Sage
    [abs(gaussian_binomial(n,1,-12)) for n in range(0,18)] # Zerinvary Lajos, May 28 2009
    

Formula

From Emeric Deutsch, Apr 01 2004: (Start)
a(n) = 12^(n-1) - a(n-1).
G.f.: x/(1 - 11*x - 12*x^2). (End)
E.g.f.: exp(-x)*(exp(13*x) - 1)/13. - Stefano Spezia, Mar 11 2020

A090409 a(n) = (7*8^n + 2*(-1)^n)/9.

Original entry on oeis.org

1, 6, 50, 398, 3186, 25486, 203890, 1631118, 13048946, 104391566, 835132530, 6681060238, 53448481906, 427587855246, 3420702841970, 27365622735758, 218924981886066, 1751399855088526, 14011198840708210, 112089590725665678, 896716725805325426, 7173733806442603406
Offset: 0

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Author

Paul Barry, Nov 29 2003

Keywords

Crossrefs

First differences of A015565.

Programs

  • Mathematica
    LinearRecurrence[{7,8},{1,6},20] (* Harvey P. Dale, Aug 15 2016 *)

Formula

a(n) = Sum_{j=0..2} Sum_{k=0..n} C(3*n+j, 3*k)/3.
a(n) = (A007613(n) + A082311(n) + A082365(n))/3.
G.f.: (-1+x)/((1+x)*(8*x-1)). - R. J. Mathar, Dec 10 2014
From Elmo R. Oliveira, Aug 18 2024: (Start)
E.g.f.: exp(-x)*(7*exp(9*x) + 2)/9.
a(n) = 7*a(n-1) + 8*a(n-2) for n > 1. (End)

Extensions

a(20)-a(21) from Elmo R. Oliveira, Aug 18 2024

A141355 The Jacobsthal sequence, dropping each third term.

Original entry on oeis.org

1, 1, 5, 11, 43, 85, 341, 683, 2731, 5461, 21845, 43691, 174763, 349525, 1398101, 2796203, 11184811, 22369621, 89478485, 178956971, 715827883, 1431655765, 5726623061, 11453246123, 45812984491, 91625968981, 366503875925, 733007751851
Offset: 0

Views

Author

Paul Curtz, Aug 03 2008

Keywords

Comments

A001045 after removal of the subsequence A132805.

Programs

  • Python
    def A141355(n): return ((1<<(n+1<<1)-(n+1>>1)-1)|1)//3 # Chai Wah Wu, Apr 19 2025

Formula

a(2n+1)-a(2n) = 6*A015565(n).
a(4n+1)=2a(4n)-1. a(4n+2)=4a(4n+1)+1. a(4n+3)=2a(4n+2)+1. a(4n+4)=4a(4n+3)-1.
a(2n)= A082311(n). a(2n+1) = A082365(n). - R. J. Mathar, Feb 23 2009
a(n)=7*a(n-2)+8*a(n-4). G.f.: (1+x-2*x^2+4*x^3)/((1-8*x^2)*(1+x^2)). - R. J. Mathar, Feb 23 2009

Extensions

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

A328824 Numerators of A113405(-n) (see the comment for details).

Original entry on oeis.org

0, 1, 1, 1, -7, -7, -7, 57, 57, 57, -455, -455, -455, 3641, 3641, 3641, -29127, -29127, -29127, 233017, 233017, 233017, -1864135, -1864135, -1864135, 14913081, 14913081, 14913081, -119304647, -119304647, -119304647
Offset: 0

Views

Author

Paul Curtz, Oct 28 2019

Keywords

Comments

Let A(n) = (2^n + (-1)^(n+1) - 2*sqrt(3)*sin((Pi*n)/3))/9. Then A(n) = A113405(n) and a(n) = numerator(A(-n)).

Crossrefs

Programs

  • Maple
    gf := x / ((1 - x)*(1 + 2*x)*(1 - 2*x + 4*x^2)): ser := series(gf, x, 36):
    seq(coeff(ser,x,n),n=0..30); # Peter Luschny, Nov 11 2019
  • Mathematica
    LinearRecurrence[{1,0,-8,8},{0,1,1,1},50] (* Paolo Xausa, Nov 13 2023 *)
  • PARI
    concat(0, Vec(x / ((1 - x)*(1 + 2*x)*(1 - 2*x + 4*x^2)) + O(x^40))) \\ Colin Barker, Nov 11 2019

Formula

From Colin Barker, Nov 11 2019: (Start)
G.f.: x / ((1 - x)*(1 + 2*x)*(1 - 2*x + 4*x^2)).
a(n) = a(n-1) - 8*a(n-3) + 8*a(n-4) for n>3. (End)

A328881 a(n+3) = 2^n - a(n), a(0)=a(2)=1, a(1)=0 for n >= 0.

Original entry on oeis.org

1, 0, 1, 0, 2, 3, 8, 14, 29, 56, 114, 227, 456, 910, 1821, 3640, 7282, 14563, 29128, 58254, 116509, 233016, 466034, 932067, 1864136, 3728270, 7456541, 14913080, 29826162, 59652323, 119304648, 238609294, 477218589, 954437176, 1908874354, 3817748707
Offset: 0

Views

Author

Paul Curtz, Oct 29 2019

Keywords

Comments

The array of a(n) and its repeated differences:
1, 0, 1, 0, 2, 3, 8, 14, ...
-1, 1, -1, 2, 1, 5, 6, 15, ...
2, -2, 3, -1, 4, 1, 9, 12, ...
-4, 5, -4, 5, -3, 8, 3, 19, ...
9, -9, 9, -8, 11, -5, 16, 5, ...
-18, 18, -17, 19, -16, 21, -11, 32, ...
36, -35, 36, -35, 37, -32, 43, -21, ...
-71, 71, -71, 72, -69, 75, -64, 85, ...
...
The recurrence is the same for every row.
From Jean-François Alcover, Nov 28 2019: (Start)
It appears that, when odd, a(n) is never a multiple of 5.
Main and 3rd upper diagonals of the difference array are A001045 (Jacobsthal numbers); first upper diagonal is negated A001045; second upper diagonal is A000079 (powers of 2); 4th upper diagonal is A062092.
(End)

Crossrefs

Programs

  • Mathematica
    a[0] = a[2] = 1; a[1] = 0; a[n_] := a[n] = 2^(n - 3) - a[n - 3]; Array[a, 36, 0] (* Amiram Eldar, Nov 06 2019 *)
  • PARI
    Vec((1 - 2*x + x^2 - x^3) / ((1 + x)*(1 - 2*x)*(1 - x + x^2)) + O(x^40)) \\ Colin Barker, Oct 29 2019

Formula

a(n+1) - 2*a(n) = period 6: repeat [-2, 1, -2, 2, -1, 2].
a(n+12) - a(n) = 455*2^n.
From Colin Barker, Oct 29 2019: (Start)
G.f.: (1 - 2*x + x^2 - x^3) / ((1 + x)*(1 - 2*x)*(1 - x + x^2)).
a(n) = 2*a(n-1) - a(n-3) + 2*a(n-4) for n>3.
(End)
a(n+2) - a(n) = A024495(n).
a(n+6) - a(n) = 7*2^n.
a(n+9) + a(n) = 57*2^n.
a(n) = A113405(n) + A092220(n+5).
9*a(n) = 2^n + 5*(-1)^n + 3*A010892(n). - R. J. Mathar, Nov 28 2019

A379530 a(n) = (A135318(3*n) + A135318(3*n+1) + A135318(3*n+2))/3.

Original entry on oeis.org

1, 3, 8, 23, 64, 185, 512, 1479, 4096, 11833, 32768, 94663, 262144, 757305, 2097152, 6058439, 16777216, 48467513, 134217728, 387740103, 1073741824, 3101920825, 8589934592, 24815366599, 68719476736, 198522932793, 549755813888, 1588183462343, 4398046511104, 12705467698745
Offset: 0

Views

Author

Paul Curtz, Dec 24 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0, 7, 0, 8}, {1, 3, 8, 23}, 30] (* Amiram Eldar, Dec 31 2024 *)

Formula

a(n) = 7*a(n-2) + 8*a(n-4) with a(0)=1, a(1)=3, a(2)=8, a(3)=23 for n >= 4.
a(2*n) = A001018(n).
a(2*n+1) = A015565(n+1) + A013730(n).
Previous Showing 21-26 of 26 results.