A386236 -id:A386236 - cp's OEIS frontend

A046737 Reduced period of A000073 mod n.

1, 4, 13, 8, 31, 52, 16, 16, 13, 124, 110, 104, 56, 16, 403, 32, 96, 52, 120, 248, 208, 220, 553, 208, 155, 56, 39, 16, 140, 1612, 331, 64, 1430, 96, 496, 104, 469, 120, 728, 496, 560, 208, 308, 440, 403, 2212, 46, 416, 112, 620, 1248, 56, 52, 156

Offset: 1

David W. Wilson

nonn

See A046738 for the period of the tribonacci numbers mod n. The ratio of the period to the reduced period is either 1 or 3. Robinson discusses the relationship between the period and the reduced period of a sequence. For the Fibonacci numbers, the analogous sequence is A001177. - T. D. Noe, Jan 14 2009

			From _T. D. Noe_, Jan 14 2009: (Start)
The tribonacci sequence (starting with 1) mod 7 has a period that repeats
  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.
The first pair of zeros occurs at the 16th term. Hence a(7)=16.
(End)

T. D. Noe, Table of n, a(n) for n=1..1000
D. W. Robinson, A note on linear recurrent sequences modulo m, Amer. Math. Monthly 73 (1966), 619-621.

Cf. A000073, A001177, A046738, A154753 (restriction to prime indices), A386236.

Improved name from T. D. Noe, Jan 14 2009

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

Original entry on oeis.org

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.

cp's OEIS Frontend

A046737 Reduced period of A000073 mod n.

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