A000057 Primes dividing all Fibonacci sequences.

2, 3, 7, 23, 43, 67, 83, 103, 127, 163, 167, 223, 227, 283, 367, 383, 443, 463, 467, 487, 503, 523, 547, 587, 607, 643, 647, 683, 727, 787, 823, 827, 863, 883, 887, 907, 947, 983, 1063, 1123, 1163, 1187, 1283, 1303, 1327, 1367, 1423, 1447, 1487, 1543, 1567, 1583

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn

Comments

Here a Fibonacci sequence is a sequence which begins with any two integers and continues using the rule s(n+2) = s(n+1) + s(n). These primes divide at least one number in each such sequence. - Don Reble, Dec 15 2006

Primes p such that the smallest positive m for which Fibonacci(m) == 0 (mod p) is m = p + 1. In other words, the n-th prime p is in this sequence iff A001602(n) = p + 1. - Max Alekseyev, Nov 23 2007

Cubre and Rouse comment that this sequence is not known to be infinite. - Charles R Greathouse IV, Jan 02 2013

Number of terms up to 10^n: 3, 7, 38, 249, 1894, 15456, 130824, 1134404, 10007875, 89562047, .... - Charles R Greathouse IV, Nov 19 2014

These are also the fixed points of sequence A213648 which gives the minimal number of 1's such that n*[n; 1,..., 1, n] = [x; ..., x], where [...] denotes simple continued fractions. - M. F. Hasler, Sep 15 2015

It appears that for n >= 2, all first differences are congruent to 0 (mod 4). - Christopher Hohl, Dec 28 2018

The comment above is equivalent to a(n) == 3 (mod 4) for n >= 2. This is indeed correct. Actually it can be proved that a(n) == 3, 7 (mod 20) for n >= 2. Let p != 2, 5 be a prime, then: A001175(p) divides (p - 1)/2 if p == 1, 9 (mod 20); p - 1 if p == 11, 19 (mod 20); (p + 1)/2 if p == 13, 17 (mod 20). So the remaining cases are p == 3, 7 (mod 20). - Jianing Song, Dec 29 2018

References

• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Christian G. Bower and T. D. Noe, Table of n, a(n) for n = 1..1000
• U. Alfred, Primes which are factors of all Fibonacci sequences, Fib. Quart., 2 (1964), 33-38.
• B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.
• D. M. Bloom, On periodicity in generalized Fibonacci sequences, Am. Math. Monthly 72 (8) (1965) 856-861.
• H. E. A. Campbell and David L. Wehlau, Zigzag polynomials, Artin's conjecture and trinomials, Finite Fields and Their Applications (2023) Vol. 89, 102198.
• Paul Cubre and Jeremy Rouse, Divisibility properties of the Fibonacci entry point, arXiv:1212.6221 [math.NT], 2012.
• Ron Knott, General Fibonacci Series.
• Rishi Kumar, Kepler Sets of Second-Order Linear Recurrence Sequences Over Q_p, arXiv:2406.05890 [math.NT], 2024. See pp. 2, 7.
• N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence).

Crossrefs

Subsequence of A064414.

Cf. A001175, A001602, A079346, A106535, A213648.

Programs

Mathematica

Select[Prime[Range[1000]], Function[p, a=0; b=1; n=1; While[b != 0, t=b; b = Mod[(a+b), p]; a=t; n++]; n>p]] (* Jean-François Alcover, Aug 05 2018, after Charles R Greathouse IV *)

PARI

select(p->my(a=0,b=1,n=1,t);while(b,t=b;b=(a+b)%p; a=t; n++); n>p, primes(1000)) \\ Charles R Greathouse IV, Jan 02 2013

PARI

is(p)=fordiv(p-1,d,if(((Mod([1,1;1,0],p))^d)[1,2]==0,return(0)));fordiv(p+1,d,if(((Mod([1,1;1,0],p))^d)[1,2]==0,return(d==p+1 && isprime(p)))) \\ Charles R Greathouse IV, Jan 02 2013

PARI

is(p)=if((p-2)%5>1, return(0)); my(f=factor(p+1)); for(i=1, #f~, if((Mod([1, 1; 1, 0], p)^((p+1)/f[i, 1]))[1, 2]==0, return(0))); isprime(p) \\ Charles R Greathouse IV, Nov 19 2014

Extensions

More terms from Don Reble, Nov 14 2006