A322906 The number of zeros in the fundamental Pisano period of the 3-Fibonacci numbers A006190 modulo n.

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

Offset: 1

Author

Jianing Song, Jan 05 2019

Keywords

• nonn

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

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).

Programs

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))

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).