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.

A158587 a(n) = 289*n^2 - 17.

Original entry on oeis.org

272, 1139, 2584, 4607, 7208, 10387, 14144, 18479, 23392, 28883, 34952, 41599, 48824, 56627, 65008, 73967, 83504, 93619, 104312, 115583, 127432, 139859, 152864, 166447, 180608, 195347, 210664, 226559, 243032, 260083, 277712, 295919, 314704, 334067, 354008, 374527
Offset: 1

Views

Author

Vincenzo Librandi, Mar 22 2009

Keywords

Comments

The identity (34*n^2 - 1)^2 - (289*n^2 - 17)*(2*n)^2 = 1 can be written as A158588(n)^2 - a(n)*A005843(n)^2 = 1.

Crossrefs

Programs

  • Magma
    I:=[272, 1139, 2584]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 15 2012
    
  • Mathematica
    LinearRecurrence[{3, -3, 1}, {272, 1139, 2584}, 50] (* Vincenzo Librandi, Feb 15 2012 *)
    289*Range[40]^2-17 (* Harvey P. Dale, Jan 30 2019 *)
  • PARI
    for(n=1, 50, print1(289*n^2-17", ")); \\ Vincenzo Librandi, Feb 15 2012

Formula

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 17*x*(-16 - 19*x + x^2)/(x-1)^3.
From Amiram Eldar, Mar 14 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(17))*Pi/sqrt(17))/34.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(17))*Pi/sqrt(17) - 1)/34. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: 17*(exp(x)*(17*x^2 + 17*x - 1) + 1).
a(n) = 17*A321180(n). (End)

Extensions

Comment rewritten by R. J. Mathar, Oct 16 2009

A158729 a(n) = 1156*n^2 - 34.

Original entry on oeis.org

1122, 4590, 10370, 18462, 28866, 41582, 56610, 73950, 93602, 115566, 139842, 166430, 195330, 226542, 260066, 295902, 334050, 374510, 417282, 462366, 509762, 559470, 611490, 665822, 722466, 781422, 842690, 906270, 972162, 1040366, 1110882, 1183710, 1258850, 1336302
Offset: 1

Views

Author

Vincenzo Librandi, Mar 25 2009

Keywords

Comments

The identity (68*n^2 - 1)^2 - (1156*n^2 - 34)*(2*n)^2 = 1 can be written as A158730(n)^2 - a(n)*A005843(n)^2 = 1.

Crossrefs

Programs

  • Magma
    I:=[1122, 4590, 10370]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 20 2012
    
  • Mathematica
    LinearRecurrence[{3, -3, 1}, {1122, 4590, 10370}, 50] (* Vincenzo Librandi, Feb 20 2012 *)
  • PARI
    for(n=1, 40, print1(1156*n^2 - 34", ")); \\ Vincenzo Librandi, Feb 20 2012

Formula

G.f.: 34*x*(-33 - 36*x + x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 22 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(34))*Pi/sqrt(34))/68.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(34))*Pi/sqrt(34) - 1)/68. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: 34*(exp(x)*(34*x^2 + 34*x - 1) + 1).
a(n) = 34*A158588(n). (End)

Extensions

Comment rewritten and formula replaced by R. J. Mathar, Oct 22 2009
Showing 1-2 of 2 results.