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 1-3 of 3 results.

A014409 Number of inequivalent ways (mod D_4) a pair of checkers can be placed on an n X n board.

Original entry on oeis.org

0, 2, 8, 21, 49, 93, 171, 278, 446, 660, 970, 1347, 1863, 2471, 3269, 4188, 5356, 6678, 8316, 10145, 12365, 14817, 17743, 20946, 24714, 28808, 33566, 38703, 44611, 50955, 58185, 65912, 74648, 83946, 94384, 105453, 117801, 130853, 145331, 160590, 177430, 195132
Offset: 1

Views

Author

Borghard, William (bogey(AT)hostare.att.com)

Keywords

References

  • Computed by Fred Hallden.

Crossrefs

Programs

  • Magma
    [(2*n^4+14*n^2-12*n-1-(-1)^n*(2*n^2-4*n-1))/32 : n in [1..60]]; // Wesley Ivan Hurt, Dec 30 2023
  • Mathematica
    LinearRecurrence[{2, 2, -6, 0, 6, -2, -2, 1}, {0, 2, 8, 21, 49, 93, 171, 278}, 40]
    CoefficientList[Series[- x (x^5 + x^4 + 3 x^3 + x^2 + 4 x + 2)/((x - 1)^5 (x + 1)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Oct 15 2013 *)
  • PARI
    a(n)=if(n%2, n^4 + 8*n^2 - 8*n - 1, n^4 + 6*n^2 - 4*n)/16  \\ Charles R Greathouse IV, Feb 09 2017
    

Formula

a(2*n) = n/2*(2*n^3 + 3*n - 1); a(2*n+1) = n/2*(2*n^3 + 4*n^2 + 7*n + 3).
a(0)=0, a(1)=2, a(2)=8, a(3)=21, a(4)=49, a(5)=93, a(6)=171, a(7)=278, a(n)=2*a(n-1)+2*a(n-2)-6*a(n-3)+0*a(n-4)+6*a(n-5)-2*a(n-6)- 2*a(n-7)+ a(n-8). - Harvey P. Dale, May 06 2012
G.f.: -x^2*(x^5+x^4+3*x^3+x^2+4*x+2) / ((x-1)^5*(x+1)^3). - Colin Barker, Jul 11 2013
From James Stein, May 22 2014: (Start)
For odd n: a(n) = (n^4 + 8*n^2 - 8*n - 1)/16;
For even n: a(n) = n*(n^3 + 6*n - 4)/16. (End)
a(n) = A054252(n, 2), n >= 0. - Wolfdieter Lang, Oct 03 2016
E.g.f.: (x*(1 + 13*x + 6*x^2 + x^3)*cosh(x) + (-1 + 3*x + 15*x^2 + 6*x^3 + x^4)*sinh(x))/16. - Stefano Spezia, Apr 14 2022
a(n) = (2*n^4+14*n^2-12*n-1-(-1)^n*(2*n^2-4*n-1))/32. - Wesley Ivan Hurt, Dec 30 2023

Extensions

More terms and formula from Hugo van der Sanden
More terms from Colin Barker, Jul 11 2013

A082966 Number of inequivalent ways (mod D_4) three checkers can be placed on an n X n board.

Original entry on oeis.org

0, 1, 16, 77, 319, 920, 2397, 5278, 10874, 20355, 36390, 61171, 99441, 154882, 235179, 346060, 499172, 702933, 974124, 1324585, 1777555, 2349116, 3070441, 3962762, 5066814, 6409975, 8044322, 10004463, 12355749, 15141190, 18441495, 22309336, 26843016, 32106217
Offset: 1

Views

Author

Vladeta Jovovic, May 27 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Rest@ CoefficientList[Series[x^2*(x^8 - x^7 - 4 x^6 - 55 x^5 - 56 x^4 - 83 x^3 - 28 x^2 - 13 x - 1)/((x - 1)^7*(x + 1)^4), {x, 0, 34}], x] (* Michael De Vlieger, Oct 03 2016 *)

Formula

a(n) = (1/48)*(n-1)*(n^5+n^4-2*n^3+14*n^2-5*n+3) if n is odd;
a(n) = (1/48)*n*(n-1)*(n^2-n+2)*(n^2+2*n-2) if n is even.
G.f.: x^2*(x^8-x^7-4*x^6-55*x^5-56*x^4-83*x^3-28*x^2-13*x-1) / ((x-1)^7*(x+1)^4). - Colin Barker, Jul 11 2013
a(n) = A054247(n, 3) = A054247(n, n^2-3), n >= 1. - Wolfdieter Lang, Oct 03 2016
E.g.f.: (x*(3 + 24*x + 88*x^2 + 62*x^3 + 15*x^4 + x^5)*cosh(x) + (-3 + 39*x^2 + 80*x^3 + 62*x^4 + 15*x^5 + x^6)*sinh(x))/48. - Stefano Spezia, Apr 14 2022

Extensions

More terms from Colin Barker, Jul 11 2013

A242358 Number of inequivalent (mod D_4) ways five checkers can be placed on an n X n board.

Original entry on oeis.org

23, 567, 6814, 47358, 239511, 954226, 3207212, 9414828, 24862239, 60136329, 135311658, 286229762, 574460495, 1101240084, 2028333848, 3605765688, 6211552455, 10402472811, 16984387958, 27099325638, 42342870823, 64905898662, 97761436356, 144885584740, 211543443215
Offset: 3

Views

Author

Heinrich Ludwig, May 11 2014

Keywords

Crossrefs

Programs

  • Mathematica
    Drop[CoefficientList[Series[x^3*(-23 - 452*x - 4071*x^2 - 16016*x^3 - 40397*x^4 - 59335*x^5 - 61954*x^6 - 38236*x^7 - 17221*x^8 - 3614*x^9 - 623*x^10 + 20*x^11 + x^12 + x^13)/((x-1)^11*(x+1)^6), {x, 0, 20}], x],3] (* Vaclav Kotesovec, May 11 2014 *)

Formula

a(n) = (n^10 - 10*n^8 + 35*n^6 + 52*n^5 - 210*n^4 + 140*n^3 - 56*n^2 + 48*n + IF(MOD(n, 2) = 1)*(52*n^5 - 145*n^4 + 140*n^3 - 80*n^2 + 48*n - 15))/960.
G.f.: x^3*(-23 - 452*x - 4071*x^2 - 16016*x^3 - 40397*x^4 - 59335*x^5 - 61954*x^6 - 38236*x^7 - 17221*x^8 - 3614*x^9 - 623*x^10 + 20*x^11 + x^12 + x^13)/((x-1)^11*(x+1)^6). - Vaclav Kotesovec, May 11 2014
a(n) = A054772(n, 5), n >=3. - Wolfdieter Lang, Oct 03 2016
Showing 1-3 of 3 results.