A182554 -id:A182554 - cp's OEIS frontend

A260228 a(n) = max(gcd(n,F(n-1)),gcd(n,F(n+1))), where F(n) is the n-th Fibonacci number.

1, 2, 3, 2, 1, 1, 7, 2, 3, 2, 11, 1, 13, 2, 3, 2, 17, 1, 19, 2, 3, 2, 23, 1, 1, 2, 3, 2, 29, 1, 31, 2, 3, 2, 1, 1, 37, 2, 3, 2, 41, 1, 43, 2, 3, 2, 47, 1, 7, 2, 3, 2, 53, 1, 1, 2, 3, 2, 59, 1, 61, 2, 21, 2, 1, 1, 67, 2, 3, 2, 71, 1, 73, 2, 3, 2, 1, 13, 79, 2, 3, 2, 83, 1, 1, 2, 3, 2, 89, 1, 1, 2, 3, 2, 1, 1, 97, 2, 33, 2

Offset: 1

Dmitry Kamenetsky, Jul 20 2015

nonn

This sequence seems very good at generating primes. Many primes p are generated when a(p)=p. In fact for n<=10000, a(n)=n occurs 1242 times and 1228 of these times n is prime. When n is a composite we often have a(n)A182554).
For n<=1000000 and a(n)=n this sequence generates all primes in the given range (except 5) and produces a prime 99.73% of the time.
It is known that if p is a prime greater than 5 then either F(p-1) or F(p+1) is a multiple of p (see Lawrence and Burton references) and so for those cases we have a(p)=p.

			a(2) = max(gcd(2,F(1)),gcd(2,F(3))) = max(1,2)=2.
a(11) = max(gcd(11,F(10)),gcd(11,F(12))) = max(gcd(11,55),gcd(11,144)) = max(11,1) = 11.
a(13) = max(gcd(13,144),gcd(13,377)) = 13.
a(23) = max(gcd(23,17711),gcd(23,46368)) = 23.

David M. Burton, Elementary Number Theory, Allyn and Bacon, p. 292, 1980.

Dmitry Kamenetsky, Table of n, a(n) for n = 1..10000
D. E. Daykin and L. A. G. Dresel, Factorization of Fibonacci numbers, Fibonacci Quarterly 8:1 (1970), pp. 23-30.
Ron Knott, Factorisation of the first 300 Fibonacci numbers
Brian Lawrence, Fibonacci numbers modulo p, Stanford SUMO talk, 2014.
Eric W. Weisstein‘s World of Mathematics, Prime Formulas
Wikipedia, Formula for primes

Cf. A106108, A182554, A260222.

[Max(Gcd(n,Fibonacci(n-1)),Gcd(n,Fibonacci(n+1))): n in [1..90]]; // Vincenzo Librandi, Jul 20 2015

Table[Max[GCD[n, Fibonacci[n - 1]], GCD[n, Fibonacci[n + 1]]], {n, 1, 80}] (* Vincenzo Librandi, Jul 20 2015 *)

first(m)=vector(m,n,max(gcd(n,fibonacci(n-1)),gcd(n,fibonacci(n+1)))) /* Anders Hellström, Jul 20 2015 */

A182504 Composite numbers k that divide both Fibonacci(k+1) and Fibonacci(2k+1)-1.

Original entry on oeis.org

323, 377, 3827, 5777, 6479, 10877, 11663, 18407, 19043, 20999, 23407, 25877, 27323, 34943, 35207, 39203, 44099, 47519, 50183, 51983, 53663, 60377, 65471, 75077, 78089, 79547, 80189, 81719, 82983, 84279, 84419, 86063, 90287, 94667, 100127, 104663, 109871

Offset: 1

Gary Detlefs, May 05 2012

nonn

A subset of A182554 based on a refinement of the Fibonacci criterion for primality described there. The additional constraint that k divides Fibonacci(2*k+1)-1 is suggested by the Cloitre comment in A003631.
What base-2 pseudoprimes are contained in this sequence?
An almost identical sequence can be obtained by testing for composite numbers for which (1) k divides Fibonacci(k+1) and (2) k^12 mod 210 = 1. All primes greater than 7 appear to satisfy condition 2. Terms of {a(n)} which are not pseudoprimes to this criterion are 50183, 65471, 82983, and 84279. - Gary Detlefs, Jun 04 2012

Giovanni Resta, Table of n, a(n) for n = 1..1000

Cf. A182554, A003631, A038872, A000040.

[n: n in [4..11*10^4] | not IsPrime(n) and IsDivisibleBy(Fibonacci(n+1),n) and IsDivisibleBy(Fibonacci(2*n+1)-1,n)]; // Bruno Berselli, May 04 2012

with (combinat): f:= n-> fibonacci(n): for n from 2 to 100000 do if not isprime(n) and irem(f(n+1), n)=0 and irem((f(2*n+1)-1), n)=0 then print(n) fi od;

Select[Range[110000],CompositeQ[#]&&Mod[Fibonacci[#+1],#]==Mod[Fibonacci[ 2#+1]-1,#] == 0&] (* Harvey P. Dale, Aug 02 2024 *)

A214434 Composite numbers k such that k divides Fibonacci(k+1) or Fibonacci(k-1) and 2^(k-1) == 1 (mod k).

Original entry on oeis.org

6601, 13981, 30889, 68101, 219781, 252601, 332949, 399001, 512461, 642001, 721801, 722261, 741751, 852841, 873181, 1024651, 1141141, 1193221, 1207361, 1533601, 1690501, 1735841, 1857241, 1909001, 2085301, 2100901, 2165801, 2603381, 2704801, 2757241, 3186821, 3568661

Offset: 1

Gary Detlefs, Jul 17 2012

nonn

Pseudoprimes to a criterion for primality which tests that
1. k divides Fibonacci(k+1) or Fibonacci(k-1) (see A182554, A081264), and
2. 2^(k-1) == 1 (mod k) (see A001567).
All terms appear to be congruent to 1 or -1 (mod 5).
Terms that are not congruent to 1 or -1 (mod 5): 22711873, 40160737, 55462177, ... . - Amiram Eldar, Sep 12 2022

			6601 is in the sequence because the 6600th Fibonacci number is divisible by 6601 and 2^6600 = 1 mod 6601.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..671 from Giovanni Resta)

Cf. A182554, A081264, A001567.

with(combinat):f:= n-> fibonacci(n): for n from 1 to 2000000 do if(f(n+1) mod n = 0 or f(n-1) mod n = 0) and 2^(n-1) mod n = 1 and not isprime(n) then print(n) fi od;

Select[Range[1, 4*10^6, 2], CompositeQ[#] && PowerMod[2, # - 1, #] == 1 && (Divisible[Fibonacci[# - 1], #] || Divisible[Fibonacci[# + 1], #]) &] (* Amiram Eldar, Sep 12 2022 *)

Intersection of A182554 and A001567.

cp's OEIS Frontend

A260228 a(n) = max(gcd(n,F(n-1)),gcd(n,F(n+1))), where F(n) is the n-th Fibonacci number.

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A182504 Composite numbers k that divide both Fibonacci(k+1) and Fibonacci(2k+1)-1.

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A214434 Composite numbers k such that k divides Fibonacci(k+1) or Fibonacci(k-1) and 2^(k-1) == 1 (mod k).

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