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A046738
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Period of Fibonacci 3-step sequence A000073 mod n.
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19
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1, 4, 13, 8, 31, 52, 48, 16, 39, 124, 110, 104, 168, 48, 403, 32, 96, 156, 360, 248, 624, 220, 553, 208, 155, 168, 117, 48, 140, 1612, 331, 64, 1430, 96, 1488, 312, 469, 360, 2184, 496, 560, 624, 308, 440, 1209, 2212, 46, 416, 336, 620, 1248, 168
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OFFSET
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1,2
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COMMENTS
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Could also be called the tribonacci Pisano periods. [Carl R. White, Oct 05 2009]
Klaska notes that n=208919=59*3541 satisfies a(n) = a(n^2). - Michel Marcus, Mar 03 2016
39, 78, 273, 546 also satisfy a(n) = a(n^2). - Michel Marcus, Mar 07 2016
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LINKS
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FORMULA
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a(3^k) = 13*3^(k-1) for k > 0. If a(p) != a(p^2) for p prime, then a(p^k) = p^(k-1)*a(p) for k > 0. [Waddill, 1978] - Chai Wah Wu, Feb 25 2022
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MAPLE
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a:= proc(n) local f, k, l; l:= ifactors(n)[2];
if nops(l)<>1 then ilcm(seq(a(i[1]^i[2]), i=l))
else f:= [0, 0, 1];
for k do f:=[f[2], f[3], f[1]+f[2]+f[3] mod n];
if f=[0, 0, 1] then break fi
od; k
fi
end:
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MATHEMATICA
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Table[a = {0, 1, 1}; a = a0 = Mod[a, n]; k = 0; While[k++; s = a[[3]] + a[[2]] + a[[1]]; a = RotateLeft[a]; a[[-1]] = Mod[s, n]; a != a0]; k, {n, 100}] (* T. D. Noe, Aug 28 2012 *)
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PROG
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(Python)
from itertools import count
a = b = (0, 0, 1%n)
for m in count(1):
b = b[1:] + (sum(b) % n, )
if a == b:
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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