A154755 -id:A154755 - cp's OEIS frontend

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

T. D. Noe, Jan 15 2009

nonn

See A046737 for more information about the reduced period.

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 1's as 3's.

			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.

See the comments for the relationships with A116515, A154755, A154756.
See the formula section for the relationships with A106302, A154753, A386236.
Cf. A000073.
For the periods modulo all positive integers see A046737, A046738.

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}]

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

a(n) = A386236(prime(n)), where prime(n) is the n-th prime.

A154756 Primes p such that ratio in A154754 is 3.

Original entry on oeis.org

7, 13, 19, 61, 73, 79, 103, 109, 127, 139, 151, 163, 193, 199, 211, 241, 271, 307, 337, 349, 373, 409, 421, 523, 541, 547, 571, 607, 613, 673, 739, 757, 769, 787, 811, 853, 877, 883, 907, 919, 937, 967, 991, 1009, 1033, 1063, 1087, 1117, 1123, 1129, 1201

Offset: 1

T. D. Noe, Jan 15 2009

nonn

Note that all these primes have the form 6k+1, which is required for the equation x^2+x+1=0 (mod p) to have two integer solutions. However, this sequence has only about half of all 6k+1 primes. What other condition determines the p in this sequence? See A154755 for the primes not in this sequence.

Cf. A000073

cp's OEIS Frontend

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

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