

A322906


The number of zeros in the fundamental Pisano period of the 3Fibonacci numbers A006190 modulo n.


11



1, 1, 1, 1, 4, 1, 2, 2, 1, 4, 2, 1, 4, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 4, 4, 1, 2, 4, 2, 2, 2, 2, 2, 2, 1, 4, 2, 2, 2, 4, 2, 1, 2, 2, 1, 2, 2, 2, 4, 2, 2, 1, 1, 2, 2, 2, 4, 2, 2, 1, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 4, 4, 2, 2, 2, 2, 1, 2, 1, 4, 2, 2, 2, 1, 2
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OFFSET

1,5


COMMENTS

a(n) is the multiplicative order of A006190(A322907(n)+1) modulo n.
a(n) has value 1, 2 or 4. This is because A006190(k,m+1)^4 == 1 (mod A006190(k,m)).
Conjecture: For primes p == 1, 9, 17, 25, 49, 81 (mod 104), the probability of a(p^e) taking on the value 1, 2, 4 is 1/6, 2/3, 1/6, respectively; for primes p == 29, 53, 61, 69, 77, 101 (mod 104), the probability of a(p^e) taking on the value 1, 4 is 1/2, 1/2, respectively.


LINKS

Jianing Song, Table of n, a(n) for n = 1..1000


FORMULA

For n > 2, T(n,k) = 4 iff A322907(n) is odd; 1 iff A322907(n) is even but not divisible by 4; 2 iff A322907(n) is divisible by 4.
For primes p == 3, 23, 27, 35, 43, 51 (mod 52), a(p^e) = 1.
For primes p == 5, 21, 33, 37, 41, 45 (mod 52), a(p^e) = 4.
For primes p == 7, 11, 15, 19, 31, 47 (mod 52), a(p^e) = 2.
a(13^e) = 4. a(2^e) = 1 if e = 1, 2 and 2 if e >= 3.
a(n) = A175182(n)/A322907(n).


PROG

(PARI) A006190(m) = ([3, 1; 1, 0]^m)[2, 1]
a(n) = my(i=1); while(A006190(i)%n!=0, i++); znorder(Mod(A006190(i+1), n))


CROSSREFS

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = k*x(n+1) + x(n). Then the periods, ranks and the ratios of the periods to the ranks modulo a given integer n are given by:
k = 1: A001175 (periods), A001177 (ranks), A001176 (ratios).
k = 2: A175181 (periods), A214028 (ranks), A214027 (ratios).
k = 3: A175182 (periods), A322907 (ranks), this sequence (ratios).
Sequence in context: A300657 A112621 A081448 * A106437 A279605 A054713
Adjacent sequences: A322903 A322904 A322905 * A322907 A322908 A322909


KEYWORD

nonn


AUTHOR

Jianing Song, Jan 05 2019


STATUS

approved



