

A236479


Factors relating prime p to the periodicity, k, for the division of p into F(k*i), the Fibonacci numbers (A000045), i >= 0.


3



1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 3, 2, 1, 4, 1, 1, 2, 1, 1, 8, 2, 2, 1, 3, 4, 6, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 1, 2, 1, 2, 2, 9, 5, 1, 1, 2, 18, 1, 2, 1, 2, 3, 4, 1, 2, 10, 1, 2, 7, 1, 2, 2, 3, 2, 3, 2, 6, 1, 1, 2, 1, 1, 4, 2, 4, 2
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OFFSET

1,6


COMMENTS

a(n) is the factor by which the periodicity of divisibility of Fibonacci numbers by the prime p is reduced relative to p itself, with adjustments on each p(n) = A000040(n) of +/ 1 or 0 as detailed below.
The periodicity, k, for the division of F(k*i) by prime(n) is given by A001602(n), for every i >= 0.
For primes p(n) congruent to {1,4} mod 5, k = (p(n)1)/a(n).
For primes p(n) congruent to {2,3} mod 5, k = (p(n)+1)/a(n).
Only at prime p(3) = 5, does k = p(3) = 5 (no reduction, no adjustment).
Thus, not only are all primes divisible into Fibonacci numbers in a simple periodic pattern, but their periods all begin at F(0)= 0, which is uncommon among other sequences in the OEIS, except for simple power sequences.


LINKS

Patrick McKinley, Table of n, a(n) for n = 1..10000


FORMULA

a(n) = (p(n)1)/A001602(n) if p is congruent to {1,4} mod 5. See A045468.
a(n) = (p(n)+1)/A001602(n) if p is congruent to {2,3} mod 5. See A003631.
a(3) = p(3)/A001602(3) = 5/5 = 1 for p = 5.


EXAMPLE

a(15) = 3, A000040(15) = 47 which is congruent with to {2,3} mod 5. Therefore the periodicity A001602(15) = k = (47+1)/3 = 16, and therefore F(16*i)/47 is always an integer.
Now consider an example of finding a(n) for the prime 1597. This prime is A000040(251) and also equals F(17), which is its first occurrence as a divisor of Fibonacci numbers. Therefore k = 17. Thus we can find a(251) as (1597+1)/17 = 94, and F(17*i)/1597, for i>= 0, is an integer.
Other F(k*i)/p examples can be found at A001076, A004187, A049666 for primes 2, 3 and 5, where k = 3, 4, and 5 respectively, but a(n) = 1.


PROG

(PARI) a(n) = {if (n==3, return (1)); p = prime(n); pm5 = p % 5; k = A001602(n); if ((pm5 == 1)  (pm5 == 4), (p1)/k, (p+1)/k); } \\ Michel Marcus, Sep 10 2014


CROSSREFS

Cf. A000045, A045468, A003631, A000040
Sequence in context: A000002 A074295 A331348 * A116514 A334028 A124767
Adjacent sequences: A236476 A236477 A236478 * A236480 A236481 A236482


KEYWORD

nonn


AUTHOR

Richard R. Forberg, Jan 26 2014


EXTENSIONS

a(78) corrected by Patrick McKinley, Aug 17 2014


STATUS

approved



