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.

A158556 a(n) = 28*n^2 + 1.

Original entry on oeis.org

1, 29, 113, 253, 449, 701, 1009, 1373, 1793, 2269, 2801, 3389, 4033, 4733, 5489, 6301, 7169, 8093, 9073, 10109, 11201, 12349, 13553, 14813, 16129, 17501, 18929, 20413, 21953, 23549, 25201, 26909, 28673, 30493, 32369, 34301, 36289, 38333, 40433, 42589, 44801, 47069
Offset: 0

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Author

Vincenzo Librandi, Mar 21 2009

Keywords

Comments

The identity (28*n^2 + 1)^2 - (196*n^2 + 14) * (2*n)^2 = 1 can be written as a(n)^2 - A158555(n)*A005843(n)^2 = 1.

Crossrefs

Programs

  • Magma
    I:=[1, 29, 113]; [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, Mar 02 2012
    
  • Mathematica
    LinearRecurrence[{3, -3, 1}, {1, 29, 113}, 50] (* Vincenzo Librandi, Mar 02 2012 *)
    28*Range[0,40]^2+1 (* Harvey P. Dale, Jun 30 2022 *)
  • PARI
    for(n=0, 40, print1(28*n^2+1", ")); \\ Vincenzo Librandi, Mar 02 2012

Formula

G.f.: (1 + 26*x + 29*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 09 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(28))*Pi/sqrt(28) + 1)/2.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(28))*Pi/sqrt(28) + 1)/2. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: exp(x)*(1 + 28*x + 28*x^2).
a(n) = A247541(2*n). (End)

Extensions

Comment rewritten, a(0) added by R. J. Mathar, Oct 16 2009

A158481 a(n) = 49*n^2 + 7.

Original entry on oeis.org

56, 203, 448, 791, 1232, 1771, 2408, 3143, 3976, 4907, 5936, 7063, 8288, 9611, 11032, 12551, 14168, 15883, 17696, 19607, 21616, 23723, 25928, 28231, 30632, 33131, 35728, 38423, 41216, 44107, 47096, 50183, 53368, 56651, 60032, 63511, 67088, 70763, 74536, 78407
Offset: 1

Views

Author

Vincenzo Librandi, Mar 20 2009

Keywords

Comments

The identity (14*n^2 + 1)^2 - (49*n^2 + 7)*(2*n)^2 = 1 can be written as A158482(n)^2 - a(n)*A005843(n)^2 = 1.

Crossrefs

Programs

  • Magma
    I:=[56, 203, 448];[n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]];
    
  • Mathematica
    LinearRecurrence[{3,-3,1},{56,203,448},40]
  • PARI
    a(n)=49*n^2+7.

Formula

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f: 7*x*(8+5*x+x^2)/(1-x)^3.
From Amiram Eldar, Mar 05 2023: (Start)
Sum_{n>=1} 1/a(n) = (coth(Pi/sqrt(7))*Pi/sqrt(7) - 1)/14.
Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/sqrt(7))*Pi/sqrt(7))/14. (End)
From Elmo R. Oliveira, Jan 15 2025: (Start)
E.g.f.: 7*(exp(x)*(7*x^2 + 7*x + 1) - 1).
a(n) = 7*A247541(n). (End)
Showing 1-2 of 2 results.