

A000057


Primes dividing all Fibonacci sequences.
(Formerly M0856 N0326)


43



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
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OFFSET

1,1


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 nth prime p is in this sequence iff A001602(n) = p + 1.  Max Alekseyev, Nov 23 2007
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



MATHEMATICA



PROG

(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(p1, 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((p2)%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


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



