A001176 - OEIS

A001176

Number of zeros in fundamental period of Fibonacci numbers mod n.
(Formerly M0165 N0064)

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,3

COMMENTS

If the Fibonacci numbers are indexed so that 3 is the fourth number, then if the modulo base is a Fibonacci number (>= 3) with an even index, the period has 2 zeros. If the base is a Fibonacci number (>= 5) with an odd index, the period has 4 zeros. - Kerry Mitchell, Dec 11 2005

For a proof that A001177(n) divides the period length A001175(n) for n >= 1, see, e.g., the Vajda reference, p. 73. This comment refers to the present first formula. - Wolfdieter Lang, Jan 19 2015

REFERENCES

B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers. Report ORNL-4261, Oak Ridge National Laboratory, Oak Ridge, Tennessee, Jun 1968.

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).

S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

LINKS

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

Brennan Benfield and Michelle Manes, The Fibonacci Sequence is Normal Base 10, arXiv:2202.08986 [math.NT], 2022.

J. D. Fulton and W. L. Morris, On arithmetical functions related to the Fibonacci numbers, Acta Arithmetica, 16 (1969), 105-110.

B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers [Annotated and scanned copy]

M. Renault, Fibonacci sequence modulo m

Review of B. H. Hannon and W. L. Morris tables, Math. Comp., 23 (1969), 459-460.

FORMULA

a(n) = A001175(n)/A001177(n) for n >= 1.

a(n) = ord(n, fibonacci(A001177(n) + 1)), where ord(n, a) is the multiplicative order of a modulo n. - Mircea Merca, Jan 03 2011

a(n) = A128924(n,1). - Reinhard Zumkeller, Jan 17 2014

From Isaac Saffold, Aug 30 2018: (Start)

With the sole exception of a(8) = 2,

a(p^k) = 1 if A007814(A001175(p^k)) < 2.

a(p^k) = 4 if A007814(A001175(p^k)) = 2.

a(p^k) = 2 if A007814(A001175(p^k)) > 2. (End)

From Jianing Song, Sep 01 2018: (Start)

a(2^e) = 1 if e <= 2, otherwise 2. For odd primes p, a(p^e) = 4 if A001177(p) is odd; 1 if A001177(p) is even but not divisible by 4; 2 if A001177(p) is divisible by 4.

a(n) = 2 for n == 0, 3, 7, 8, 12, 15 (mod 20). a(p^e) = 1 if primes p == 11, 19 (mod 20); 4 if p == 13, 17 (mod 20). Conjecture: 1/6 of the primes congruent to 1 or 9 mod 40 satisfy a(p^e) = 1, 2/3 of them satisfy a(p^e) = 2 and 1/6 of them satisfy a(p^e) = 4; also, 1/2 of the primes congruent to 21 or 29 mod 40 satisfy a(p^e) = 1 and 1/2 of them satisfy a(p^e) = 4. (End)

EXAMPLE

{F(n) mod 1} has fundamental period (0) with 1 zero.

{F(n) mod 2} has fundamental period (0,1,1) with 1 zero.

{F(n) mod 3} has fundamental period (0,1,1,2,0,2,2,1) with 2 zeros.

{F(n) mod 4} has fundamental period (0,1,1,2,3,1), with 1 zero.

{F(n) mod 5} has fundamental period (0,1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1) with 4 zeros.

MATHEMATICA

With[{fibs=Fibonacci[Range[2000]]}, Table[Count[FindTransientRepeat[ Mod[ fibs, n], 3][[2]], 0], {n, 110}]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 26 2016 *)

PROG

(Haskell)

a001176 1 = 1

a001176 n = f 1 ps 0 where

f 0 (1 : xs) z = z

f _ (x : xs) z = f x xs (z + 0 ^ x)

ps = 1 : 1 : zipWith (\u v -> (u + v) `mod` n) (tail ps) ps

-- Reinhard Zumkeller, Jan 15 2014

CROSSREFS

Cf. A001175, A001177, A053027, A053028, A053029, A053030, A053031, A053032.

Cf. A235715.

Sequence in context: A365425 A365381 A161822 * A136693 A348045 A086685