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 1-5 of 5 results.

A125253 Primes p that divide Fibonacci[(p-1)/7].

Original entry on oeis.org

2269, 2521, 2731, 2969, 3571, 3739, 4481, 4831, 5741, 6091, 6329, 6581, 9521, 10949, 11159, 11789, 12391, 13049, 13679, 14281, 14449, 14771, 16829, 16871, 18229, 19489, 19559, 20021, 20399, 21701, 23269, 24179, 24571, 26111, 29191, 31039
Offset: 1

Views

Author

Alexander Adamchuk, Nov 26 2006

Keywords

Links

Crossrefs

Cf. A125252 = Primes p that divide Fibonacci[(p+1)/7]. Cf. A122487 = Primes p that divide Fibonacci[(p+1)/2]. Cf. A047652 = Primes p that divide Fibonacci[(p-1)/3]. Cf. A001583 = Artiads. Primes p that divide Fibonacci[(p-1)/5].

Programs

  • Mathematica
    Select[Prime[Range[5000]], IntegerQ[Fibonacci[(#1-1)/7]/#1]&]

A168171 Least prime p = 1 (mod n) which divides Fibonacci((p-1)/n).

Original entry on oeis.org

11, 29, 139, 61, 211, 541, 2269, 89, 199, 281, 859, 661, 911, 2269, 2221, 2081, 2789, 2161, 3041, 421, 2521, 19009, 21529, 3001, 9901, 5981, 2161, 2269, 26449, 2221, 31249, 19681, 17491, 2789, 3571, 25309, 30859, 3041, 6709, 3001, 9349, 2521, 13159, 19009
Offset: 1

Views

Author

M. F. Hasler, Nov 25 2009

Keywords

Examples

			For n=1, all numbers p satisfy p=1 (mod n), but p=11 is the least prime that divides F((p-1)/1)=F(p-1)=F(10)=55.
For n=2, all odd numbers, thus all primes p>2, satisfy p=1 (mod n), but p=29 is the first one to divide F((p-1)/2) = F(14) = 377 = 13*29.
For n=5, a(n)=211 is the smallest Artiad, i.e. prime p=1 (mod 5) which divides F((p-1)/5) = F(42) = 211*1269736.

Links

Crossrefs

Cf. A122487 (p | F[(p+1)/2]), A047652 (p | F[(p-1)/3]), A001583 (Artiads: p | F[(p-1)/5]), A125252 (p | F[(p+1)/7]), A125253 (p | F[(p-1)/7]).

Programs

  • Mathematica
    a[1] = 11;
a[n_] := For[p = 1, True, p = p + n, If[PrimeQ[p] && Divisible[Fibonacci[(p - 1)/n], p], Return[p]]];
a /@ Range[100] (* Jean-François Alcover, Oct 14 2019 *)
  • PARI
    for(n=1,99,forprime(p=1,oo,(p-1)%n & next; fibonacci((p-1)/n)%p || print1(p, ", ") || next(2)))

A124096 Primes p that divide Fibonacci[(p+1)/3].

Original entry on oeis.org

47, 107, 113, 233, 263, 347, 353, 557, 563, 677, 743, 953, 977, 1097, 1217, 1223, 1277, 1307, 1523, 1553, 1733, 1823, 1877, 1913, 1973, 2027, 2207, 2237, 2243, 2267, 2333, 2417, 2447, 2663, 2687, 2753, 2777, 3167, 3323, 3347, 3407, 3467, 3533, 3557, 3617
Offset: 1

Views

Author

Alexander Adamchuk, Nov 26 2006

Keywords

Links

Crossrefs

Cf. A047652 = Primes p that divide Fibonacci[(p-1)/3]. Cf. A122487 = Primes p that divide Fibonacci[(p+1)/2].

Programs

  • Mathematica
    Select[Prime[Range[1000]], IntegerQ[Fibonacci[(#1+1)/3]/#1]&]

A168172 Least prime p == -1 (mod n) that divides Fibonacci((p+1)/n), or 0 if no such prime exists.

Original entry on oeis.org

2, 13, 47, 0, 0, 113, 307, 0, 233, 0, 967, 0, 2417, 797, 0, 0, 1087, 233, 5737, 0, 5417, 5653, 1103, 0, 0, 2417, 4373, 0, 6263, 0, 25357, 0, 3167, 42533, 0, 0, 4513, 5737, 2417, 0, 61417, 5417, 32507, 0, 0, 36017, 1597, 0, 97607, 0, 27947, 0, 42293, 4373, 0, 0
Offset: 1

Views

Author

M. F. Hasler, Nov 28 2009

Keywords

Comments

Max Alekseyev has proved (cf. link) that a(n)=0 if n is a multiple of 4 or 5; for all other n, a prime a(n) with the required property seems to exist.

Links

Crossrefs

Cf. A168171 (least p | F[(p-1)/n]), A122487 (p | F[(p+1)/2]), A047652 (p | F[(p-1)/3]), A001583 (Artiads: p | F[(p-1)/5]), A125252 (p | F[(p+1)/7]), A125253 (p | F[(p-1)/7]).

Programs

  • PARI
    A168172(n) = n%4 && n%5 && forstep(p=n-1,1e9,n, isprime(p) || next; fibonacci((p+1)/n)%p || return(p))

A366951 a(n) = 2*(p_n - 1)/A060305(n) iff p_n == +/- 1 (mod 5), 2*(p_n + 1)/A060305(n) iff p_n == +/- 2 (mod 5), 0 iff p_n = 5.

Original entry on oeis.org

2, 1, 0, 1, 2, 1, 1, 2, 1, 4, 2, 1, 2, 1, 3, 1, 2, 2, 1, 2, 1, 2, 1, 4, 1, 4, 1, 3, 2, 3, 1, 2, 1, 6, 2, 6, 1, 1, 1, 1, 2, 4, 2, 1, 1, 18, 10, 1, 1, 4, 9, 2, 2, 2, 1, 3, 2, 2, 1, 10, 1, 1, 7, 2, 1, 1, 6, 1, 3, 4, 3, 2, 1, 1, 2, 1, 2, 1, 4, 2, 2, 10, 2, 1, 2, 1
Offset: 1

Views

Author

A.H.M. Smeets, Oct 29 2023

Keywords

Crossrefs

Cf. A000040, A003147, A060305, A047650, A047652, A071774, A124096, A124253, A296240, A308784 - A308794.

Formula

a(n) == 0 (mod 2) for prime(n) == +/- 1 (mod 5) and n > 2.
a(n) == 1 (mod 2) for Prime(n) == +/- 2 (mod 5) and n > 2.
a(n) = 1 iff prime(n) in A071774.
a(n) = 2 iff prime(n) in ({2} union A003147)/{5}.
a(n) = 3 iff prime(n) in A308784.
a(n) = 4 iff prime(n) in A308787.
a(n) = 6 iff prime(n) in A308788.
a(n) = 7 iff prime(n) in A308785.
a(n) = 8 iff prime(n) in A308789.
a(n) = 9 iff prime(n) in A308786.
a(n) = 10 iff prime(n) in A308790.
a(n) = 12 iff prime(n) in A308791.
a(n) = 14 iff prime(n) in A308792.
a(n) = 16 iff prime(n) in A308793.
a(n) = 18 iff prime(n) in A308794.
a(n) = A296240(n) iff prime(n) == +/- 2 (mod 5) and n > 3.
a(n) = 2*A296240(n) iff prime(n) == +/- 1 (mod 5) and n > 3.
a(n) in {2^k: k > 1} iff prime(n) in {A047650}.
a(n) == 3 (mod 6) iff prime(n) in {A124096}.
a(n) == 6 (mod 12) iff prime(n) in {A046652}.
a(n) == 0 (mod 14) iff prime(n) in {A125252}.
Showing 1-5 of 5 results.

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

Content is available under The OEIS End-User License Agreement.