A154754 Ratio of the period and the reduced period of the Fibonacci 3-step sequence A000073 mod prime(n).

1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 1, 1, 1, 3

Offset

1,4

Comments

For the Fibonacci 3-step (tribonacci) sequence, only 1 and 3 appear. A116515 is the analogous sequence for Fibonacci numbers. Let the terms in the reduced period be denoted by R. When the ratio is 3, the full period can be written as R,aR,bR, where a and b are multipliers that are the two solutions of the equation x^2+x+1 = 0 (mod p). What order do the solutions appear as a and b? See A154755 and A154756 for the primes that produce ratios of 1 and 3, respectively. Observe that there are approximately three times as many 1s as 3s.

Links

Table of n, a(n) for n=1..105.

Formula

a(n) = A106302(n) / A154753(n)

Example

The tribonacci sequence (starting with 1) mod 7 is 1,1,2,4,0,6,3,2,4, 2,1,0,3,4,0,0,4,4,1,2,0,3,5,1,2,1,4,0,5,2,0,0,2,2,4,1,0,5,6,4,1,4,2,0, 6,1,0,0, which has 3 pairs of 0-0 terms. Hence a(4)=3.

Mathematica

Table[p=Prime[i]; a={1, 0, 0}; a0=a; k=0; zeros=0; While[k++; s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[ -1]]=s; If[Rest[a]=={0, 0}, zeros++ ]; a!=a0]; zeros, {i, 200}]

Crossrefs

Cf. A046737, A046738.

Sequence in context: A138291 A201681 A062174 * A102368 A277109 A359262

Adjacent sequences: A154751 A154752 A154753 * A154755 A154756 A154757

Keyword

nonn

Author

T. D. Noe, Jan 15 2009

Status

approved