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.

A158657 a(n) = 784*n^2 - 28.

Original entry on oeis.org

756, 3108, 7028, 12516, 19572, 28196, 38388, 50148, 63476, 78372, 94836, 112868, 132468, 153636, 176372, 200676, 226548, 253988, 282996, 313572, 345716, 379428, 414708, 451556, 489972, 529956, 571508, 614628, 659316, 705572, 753396, 802788, 853748, 906276, 960372
Offset: 1

Views

Author

Vincenzo Librandi, Mar 23 2009

Keywords

Comments

The identity (56*n^2 - 1)^2 - (784*n^2 - 28)*(2*n)^2 = 1 can be written as A158658(n)^2 - a(n)*A005843(n)^2 = 1.

Crossrefs

Programs

  • Magma
    I:=[756, 3108, 7028]; [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 17 2012
    
  • Mathematica
    LinearRecurrence[{3, -3, 1}, {756, 3108, 7028}, 50] (* Vincenzo Librandi, Feb 17 2012 *)
  • PARI
    for(n=1, 40, print1(784*n^2 - 28", ")); \\ Vincenzo Librandi, Feb 17 2012

Formula

G.f.: 28*x*(-27 - 30*x + x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 20 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(7)))*Pi/(2*sqrt(7)))/56.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(7)))*Pi/(2*sqrt(7)) - 1)/56. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: 28*(exp(x)*(28*x^2 + 28*x - 1) + 1).
a(n) = 28*A158554(n). (End)

Extensions

Comment rephrased and redundant formula replaced by R. J. Mathar, Oct 19 2009

A158553 a(n) = 196*n^2 - 14.

Original entry on oeis.org

182, 770, 1750, 3122, 4886, 7042, 9590, 12530, 15862, 19586, 23702, 28210, 33110, 38402, 44086, 50162, 56630, 63490, 70742, 78386, 86422, 94850, 103670, 112882, 122486, 132482, 142870, 153650, 164822, 176386, 188342, 200690, 213430, 226562, 240086, 254002, 268310
Offset: 1

Views

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 A158554(n)^2 - a(n)*A005843(n)^2 = 1.

Crossrefs

Programs

  • Magma
    I:=[182, 770, 1750]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 14 2012
    
  • Mathematica
    LinearRecurrence[{3, -3, 1}, {182, 770, 1750}, 40] (* Vincenzo Librandi, Feb 14 2012 *)
    196*Range[40]^2-14 (* Harvey P. Dale, Oct 11 2023 *)
  • PARI
    for(n=1, 40, print1(196*n^2 - 14", ")); \\ Vincenzo Librandi, Feb 14 2012

Formula

G.f.: 14*x*(13 + 16*x - 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>=1} 1/a(n) = (1 - cot(Pi/sqrt(14))*Pi/sqrt(14))/28.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(14))*Pi/sqrt(14) - 1)/28. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: 14*(exp(x)*(14*x^2 + 14*x - 1) + 1).
a(n) = 14*A158485(n). (End)

Extensions

Comment rewritten by R. J. Mathar, Oct 16 2009
Showing 1-2 of 2 results.