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 A154754 Ratio of the period and the reduced period of the Fibonacci 3-step sequence A000073 mod prime(n). 3
 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 (list; graph; refs; listen; history; text; internal format)
 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 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 A063062 Adjacent sequences:  A154751 A154752 A154753 * A154755 A154756 A154757 KEYWORD nonn AUTHOR T. D. Noe, Jan 15 2009 STATUS approved

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Last modified July 24 06:00 EDT 2021. Contains 346273 sequences. (Running on oeis4.)