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A000057
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Primes dividing all Fibonacci sequences.
(Formerly M0856 N0326)
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42
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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
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1,1
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COMMENTS
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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
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
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REFERENCES
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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).
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LINKS
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MATHEMATICA
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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