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 1-4 of 4 results.

A158492 a(n) = 100*n^2 + 10.

Original entry on oeis.org

10, 110, 410, 910, 1610, 2510, 3610, 4910, 6410, 8110, 10010, 12110, 14410, 16910, 19610, 22510, 25610, 28910, 32410, 36110, 40010, 44110, 48410, 52910, 57610, 62510, 67610, 72910, 78410, 84110, 90010, 96110, 102410, 108910, 115610, 122510, 129610, 136910, 144410
Offset: 0

Views

Author

Vincenzo Librandi, Mar 21 2009

Keywords

Comments

The identity (20*n^2 + 1)^2 - (100*n^2 + 10)*(2*n)^2 = 1 can be written as A158493(n)^2 - a(n)*A005843(n)^2 = 1. - Vincenzo Librandi, Feb 21 2012

Links

Crossrefs

Cf. A005843, A158187, A158493.

Programs

  • Magma
    I:=[10, 110, 410]; [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 21 2012
  • Mathematica
    LinearRecurrence[{3, -3, 1}, {10, 110, 410}, 50] (* Vincenzo Librandi, Feb 21 2012 *)
100Range[0,40]^2+10 (* Harvey P. Dale, Dec 30 2019 *)
  • PARI
    for(n=0, 40, print1(100*n^2 + 10", ")); \\ Vincenzo Librandi, Feb 21 2012

Formula

From Vincenzo Librandi, Feb 21 2012: (Start)
G.f.: -(10 + 80*x + 110*x^2)/(x-1)^3;
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
From Amiram Eldar, Mar 05 2023: (Start)
Sum_{n>=0} 1/a(n) = (1 + coth(Pi/sqrt(10))*Pi/sqrt(10))/20.
Sum_{n>=0} (-1)^n/a(n) = (1 + cosech(Pi/sqrt(10))*Pi/sqrt(10))/20. (End)
From Elmo R. Oliveira, Jan 17 2025: (Start)
E.g.f.: 10*exp(x)*(1 + 10*x + 10*x^2).
a(n) = 10*A158187(n). (End)

Extensions

Edited by N. J. A. Sloane, Oct 12 2009

A158601 a(n) = 400*n^2 + 20.

Original entry on oeis.org

20, 420, 1620, 3620, 6420, 10020, 14420, 19620, 25620, 32420, 40020, 48420, 57620, 67620, 78420, 90020, 102420, 115620, 129620, 144420, 160020, 176420, 193620, 211620, 230420, 250020, 270420, 291620, 313620, 336420, 360020, 384420, 409620, 435620, 462420, 490020
Offset: 0

Views

Author

Vincenzo Librandi, Mar 22 2009

Keywords

Comments

The identity (40*n^2 + 1)^2 - (400*n^2 + 20)*(2*n)^2 = 1 can be written as A158602(n)^2 - a(n)*A005843(n)^2 = 1.

Links

Crossrefs

Cf. A005843, A158493, A158602.

Programs

  • Magma
    I:=[20, 420, 1620]; [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 16 2012
  • Mathematica
    400 Range[0,40]^2+20  (* Harvey P. Dale, Feb 05 2011 *)
LinearRecurrence[{3, -3, 1}, {20, 420, 1620}, 50] (* Vincenzo Librandi, Feb 16 2012 *)
  • PARI
    for(n=0, 40, print1(400*n^2 + 20", ")); \\ Vincenzo Librandi, Feb 16 2012

Formula

G.f.: -20*(1 + 18*x + 21*x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 16 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/40.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/40. (End)
From Elmo R. Oliveira, Jan 15 2025: (Start)
E.g.f.: 20*exp(x)*(1 + 20*x + 20*x^2).
a(n) = 20*A158493(n). (End)

Extensions

Comment rewritten, formula replaced by R. J. Mathar, Oct 28 2009

A158776 a(n) = 80*n^2 + 1.

Original entry on oeis.org

1, 81, 321, 721, 1281, 2001, 2881, 3921, 5121, 6481, 8001, 9681, 11521, 13521, 15681, 18001, 20481, 23121, 25921, 28881, 32001, 35281, 38721, 42321, 46081, 50001, 54081, 58321, 62721, 67281, 72001, 76881, 81921, 87121, 92481, 98001, 103681, 109521, 115521, 121681
Offset: 0

Views

Author

Vincenzo Librandi, Mar 26 2009

Keywords

Comments

The identity (80*n^2 + 1)^2 - (1600*n^2 + 40)*(2*n)^2 = 1 can be written as a(n)^2 - A158775(n)*A005843(n)^2 = 1.

Links

Crossrefs

Cf. A005843, A158493, A158775.

Programs

Formula

G.f.: -(1 + 78*x + 81*x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 24 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/(4*sqrt(5)))*Pi/(4*sqrt(5)) + 1)/2.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(4*sqrt(5)))*Pi/(4*sqrt(5)) + 1)/2. (End)
From Elmo R. Oliveira, Jan 25 2025: (Start)
E.g.f.: exp(x)*(1 + 80*x + 80*x^2).
a(n) = A158493(2*n). (End)

Extensions

Comment rewritten, a(0) added and formula replaced by R. J. Mathar, Oct 22 2009

A010022 a(0) = 1, a(n) = 40*n^2 + 2 for n>0.

Original entry on oeis.org

1, 42, 162, 362, 642, 1002, 1442, 1962, 2562, 3242, 4002, 4842, 5762, 6762, 7842, 9002, 10242, 11562, 12962, 14442, 16002, 17642, 19362, 21162, 23042, 25002, 27042, 29162, 31362, 33642, 36002, 38442, 40962, 43562, 46242, 49002, 51842, 54762, 57762, 60842
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

First bisection of A005901. - Bruno Berselli, Feb 07 2012

Links

Crossrefs

Cf. A206399.

Programs

  • Magma
    [1] cat [40*n^2+2: n in [1..50]]; // Vincenzo Librandi, Aug 03 2015
  • Mathematica
    Join[{1}, 40 Range[39]^2 + 2] (* Bruno Berselli, Feb 07 2012 *)
Join[{1}, LinearRecurrence[{3, -3, 1}, {42, 162, 362}, 50]] (* Vincenzo Librandi, Aug 03 2015 *)

Formula

G.f.: (1+x)*(1+38*x+x^2)/(1-x)^3; a(n) = A008253(4n). - Bruno Berselli, Feb 07 2012
E.g.f.: (x*(x+1)*40+2)*e^x-1. - Gopinath A. R., Feb 14 2012
Sum_{n>=0} 1/a(n) = 3/4 + sqrt(5)/40*Pi*coth(Pi*sqrt(5)/10) = 1.03983104279172.. - R. J. Mathar, May 07 2024
a(n) = 2*A158493(n), n>0. - R. J. Mathar, May 07 2024
a(n) = A195317(n)+A195317(n+1) = 2+10*A016742(n), n>0. - R. J. Mathar, May 07 2024
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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