A233281 Numbers n such that the least Fibonacci number F_k which is a multiple of n has a prime index, i.e., k is in A000040.

2, 5, 13, 37, 73, 89, 113, 149, 157, 193, 233, 269, 277, 313, 353, 389, 397, 457, 557, 613, 673, 677, 733, 757, 877, 953, 977, 997, 1069, 1093, 1153, 1213, 1237, 1453, 1597, 1657, 1753, 1873, 1877, 1933, 1949, 1993, 2017, 2137, 2221, 2237, 2309, 2333, 2417, 2473

Offset: 1

Author

Antti Karttunen, Dec 13 2013

Keywords

• nonn

Comments

Numbers n such that A001177(n) is prime.

Each natural number n belongs to this sequence if the smallest Fibonacci number which it divides is a term of A030426. - Jon E. Schoenfield, Feb 28 2014

A092395 gives all the primes in this sequence (cf. Wikipedia-link), and the first composite occurs as the 69th term, where a(69)=4181 while A092395(69)=4273. After 4181 (= 37*113 = F_19), the next term missing from A092395 is a(148)=10877 (= 73*149. A001177(10877) = 37, F_37 = 24157817 = 2221*10877). Both of these numbers (4181 and 10877) occur in various lists of Fibonacci-related pseudoprimes. Sequence A238082 gives all composites occurring in this sequence.

If n is in this sequence then all divisors d > 1 of n are in this sequence. - Charles R Greathouse IV, Feb 04 2014

Composite members begin 4181, 10877, 75077, 162133, 330929, .... - Charles R Greathouse IV, Mar 07 2014

Links

• Antti Karttunen and Charles R Greathouse IV, Table of n, a(n) for n = 1..2000 (first 157 terms from Karttunen)
• Wikipedia, Fibonacci prime, section: Divisibility of Fibonacci numbers

Crossrefs

Disjoint union of A092395 and A238082. The first 68 terms are identical with A092395, after which follows the first case of the latter sequence, with a(69) = A238082(1) = 4181.

Cf. A000045, A001177, A001602, A030426, A051694, A060442, A086597, A233282.

Programs

Haskell

a233281 n = a233281_list !! (n-1)
a233281_list = filter ((== 1) . a010051 . a001177) [1..]
-- Reinhard Zumkeller, Apr 04 2014

PARI

is(n)=my(k); while(fibonacci(k++)%n, ); isprime(k) \\ Charles R Greathouse IV, Feb 04 2014

PARI

entry(p)=my(k=1);while(fibonacci(k++)%p,);k;
is(n)={
    if(n%2==0,return(n==2));
    if(n<13, return(n==5));
    my(f=factor(n),p,F);
    if(f[1,2]>1 && f[1,1]<1e14,return(0));
    p=entry(f[1,1]);
    F=fibonacci(p);
    if(f[1,2]>1 && F%f[1,1]^f[1,2],return(0));
    if(!isprime(p), return(0));
    for(i=2,#f~,
        if(F%f[i,1]^f[i,2],return(0))
    );
    1
}; \\ Charles R Greathouse IV, Feb 04 2014

Formula

A010051(A001177(a(n))) = 1. - Reinhard Zumkeller, Apr 04 2014