A214027 The number of zeros in the fundamental Pisano period of the sequence A000129 mod n.

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

Offset: 1

Author

Art DuPre, Jul 04 2012

Keywords

• nonn

Comments

This is intimately connected with A175181 and A214028, much as A001176 is intimately connected with A001175 and A001177. In fact, A175181(n)/a(n) = A214028(n). This is the same divisibility relation that holds between A001175, A001176 and A001177.

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A175181, A214028.

Similar sequences: A001176, A322906.

Programs

Mathematica

Join[{1}, Table[s = t = Mod[{0, 1}, n]; zeros = 0; While[tmp = Mod[2*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s != t, If[tmp == 0, zeros++]]; zeros, {n, 2, 100}]] (* T. D. Noe, Jul 09 2012 *)

PARI

A000129(m) = ([2, 1; 1, 0]^m)[2, 1]
a(n) = my(i=1); while(A000129(i)%n!=0, i++); znorder(Mod(A000129(i+1), n)) \\ Jianing Song, Aug 10 2019

Formula

From Jianing Song, Sep 12 2018: (Start)

For odd primes p, a(p^e) = 4 if A214028(p) is odd; 1 if A214028(p) is even but not divisible by 4; 2 if A214028(p) is divisible by 4.

a(n) = 2 for n == 3 (mod 8). For primes p, a(p^e) = 1 if p == 7 (mod 8), 4 if p == 5 (mod 8). Conjecture: 1/6 of the primes congruent to 1 mod 8 satisfy a(p^e) = 1, 2/3 of them satisfy a(p^e) = 2 and 1/6 of them satisfy a(p^e) = 4.

(End)