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.

A271309 Integers k such that A264137(k) < A060385(k).

Original entry on oeis.org

52, 60, 65, 74, 75, 76, 85, 108, 111, 121, 124, 125, 127, 131, 132, 140, 144, 150, 153, 156, 158, 172, 175, 180, 183, 185, 195, 201, 209, 213, 216, 220, 225, 250, 263, 287, 300, 301, 327, 328, 335, 337, 339, 344, 356, 370, 402, 404, 408, 412, 417, 423, 433, 435
Offset: 1

Views

Author

Altug Alkan, Apr 03 2016

Keywords

Comments

For all corresponding values of k, A000129(k) is a composite number. In other words, k cannot be a term of A096650.

Examples

			52 is a term because A264137(52) = 66923 < A060385(52) = 90481.

Links

Crossrefs

Cf. A060385, A096650, A264137.

Programs

  • PARI
    a060385(n) = my(f=factor(fibonacci(n))[, 1]); f[#f];
a000129(n) = ([2, 1; 1, 0]^n)[2, 1];
a264137(n) = my(p=factor(a000129(n))[, 1]); p[#p];
lista(nn) = for(n=3, nn, if(a264137(n) < a060385(n), print1(n, ", ")));

Extensions

a(37)-a(54) from Amiram Eldar, May 19 2024

A246556 a(n) = smallest prime which divides Pell(n) = A000129(n) but does not divide any Pell(k) for k

Original entry on oeis.org

2, 5, 3, 29, 7, 13, 17, 197, 41, 5741, 11, 33461, 239, 269, 577, 137, 199, 37, 19, 45697, 23, 229, 1153, 1549, 79, 53, 113, 44560482149, 31, 61, 665857, 52734529, 103, 1800193921, 73, 593, 9369319, 389, 241, 1746860020068409, 4663, 11437, 43, 6481, 47, 3761, 97, 293, 45245801, 101, 22307, 68480406462161287469, 7761799, 109, 1535466241
Offset: 2

Views

Author

Eric Chen, Nov 15 2014

Keywords

Comments

First differs from A264137 (Largest prime factor of the n-th Pell number) at n=17; see Example section. - Jon E. Schoenfield, Dec 10 2016

Examples

			a(2) = 2 because Pell(2) = 2 and Pell(k) < 2 for k < 2.
a(4) = 3 because Pell(4) = 12 = 2^2 * 3, but 2 is not a primitive prime factor since Pell(2) = 2, so therefore 3 is the primitive prime factor.
a(5) = 29 because Pell(5) = 29, which is prime.
a(6) = 7 because Pell(6) = 70 = 2 * 5 * 7, but neither 2 nor 5 is a primitive prime factor, so therefore 7 is the primitive prime factor.
a(17) = 137 because Pell(17) = 1136689 = 137 * 8297, and both of them are primitive factors, we choose the smallest. (Pell(17) is the smallest Pell number with more than one primitive prime factor.)

Links

Crossrefs

Cf. A001578 (for Fibonacci(n)), A000129 (Pell numbers), A008555, A086383, A096650, A120947, A175181, A214028, A264137.

Programs

  • Mathematica
    prms={}; Table[f=First/@FactorInteger[Pell[n]]; p=Complement[f, prms]; prms=Join[prms, p]; If[p=={}, 1, First[p]], {n, 36}]

Formula

a(n) >= 2 for all n >= 2, by Carmichael's theorem. - Jonathan Sondow, Dec 08 2017

Extensions

Edited by N. J. A. Sloane, Nov 29 2014
Terms up to a(612) in b-file added by Sean A. Irvine, Sep 23 2019
Terms a(613)-a(630) in b-file added by Max Alekseyev, Aug 26 2021

A271314 Largest prime factor of the n-th Jacobsthal number, A001045(n).

Original entry on oeis.org

3, 5, 11, 7, 43, 17, 19, 31, 683, 13, 2731, 127, 331, 257, 43691, 73, 174763, 41, 5419, 683, 2796203, 241, 4051, 8191, 87211, 127, 3033169, 331, 715827883, 65537, 20857, 131071, 86171, 109, 25781083, 524287, 22366891, 61681, 8831418697, 5419, 2932031007403, 2113, 18837001
Offset: 3

Views

Author

Altug Alkan, Apr 03 2016

Keywords

Comments

a(22) = 683 is the first repeated term in this sequence. Note that a(n+2) = A129738(n), for n < 20.

Examples

			a(6) = 7 because A001045(6) = 21 = 3*7.

Links

Crossrefs

Essentially a combination of A005420 and A002587.
Cf. A001045, A006530, A060385, A129738, A264137.

Programs

  • Mathematica
    FactorInteger[#][[-1, 1]] & /@ Take[#, -(Length@ # - 3)] &@ CoefficientList[Series[x/(1 - x - 2 x^2), {x, 0, 45}], x] (* Michael De Vlieger, Apr 04 2016, after Robert G. Wilson v at A001045 *)
  • PARI
    a001045(n) = (2^n - (-1)^n) / 3;
a(n) = vecmax(factor(a001045(n))[,1]);

A364820 a(n) is the smallest prime factor of the n-th Pell number A000129(n).

Original entry on oeis.org

2, 5, 2, 29, 2, 13, 2, 5, 2, 5741, 2, 33461, 2, 5, 2, 137, 2, 37, 2, 5, 2, 229, 2, 29, 2, 5, 2, 44560482149, 2, 61, 2, 5, 2, 13, 2, 593, 2, 5, 2, 1746860020068409, 2, 11437, 2, 5, 2, 3761, 2, 13, 2, 5, 2, 68480406462161287469, 2, 29, 2, 5, 2
Offset: 2

Views

Author

Sean A. Irvine, Oct 21 2023

Keywords

Links

Crossrefs

Cf. A246556 (smallest primitive factor), A264137, A280104, A060383, A020639.

Programs

  • Mathematica
    FactorInteger[#][[1, 1]] & /@ LinearRecurrence[{2, 1}, {2, 5}, 57] (* Amiram Eldar, Oct 21 2023 *)

Formula

a(n) = A020639(A000129(n)).
Showing 1-4 of 4 results.

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

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