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A337212
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Modulo 3 Pisano period of 'n-bonacci' series.
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2
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1, 8, 13, 26, 104, 728, 364, 80, 91, 8744, 3851, 3280, 59048, 4782968, 7174453, 3438578, 16139240, 5373368, 5235412, 1678822106, 86049704, 387420488, 47071589413, 140633637386, 2952400, 728, 757, 9526168288, 7312949144072, 49566102697280, 24477226494760
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OFFSET
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1,2
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COMMENTS
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The modulo 2 variant of this sequence gives 1, 3, 4, 5, 6, 7, 8, ... (the natural numbers not including 2), and likewise, when the modulus is a power of 2, it seems that the Pisano period lengths form an arithmetic progression. (Note that both of these observations are based on empirical observation only).
a(39)=797161, a(80)=6560, a(81)=6643, a(90)=5380840, a(242)=59048, a(243)=59293, a(728)=531440, a(729)= 532171, a(2186)=4782968, a(2187)=4785157, a(6560)=43046720, a(6561)=43053283, a(19682)=387420488, a(19683)=387440173. - Chai Wah Wu, Sep 15 2020
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LINKS
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FORMULA
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Conjecture: a(3^k-1)=a(3^k)-3^k-2=3^(2k)-1, a(3^k)=3^k(3^k+1)+1 for k>0. - Chai Wah Wu, Sep 15 2020
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EXAMPLE
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For n = 3, the remainders modulo 3 of the tribonacci series are 0, 1, 1, 2, 1, 1, 1, 0, 2, 0, 2, 1, 0, (these repeat indefinitely), so the Pisano period of the 'tribonacci' sequence is 13.
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PROG
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(PARI) a(n) = {my(v=w=concat(0, vector(n-1, i, 1))); for(k=1, oo, v=concat(v[2..n], vecsum(v)%3); if(v==w, return(k))); } \\ Jinyuan Wang, Aug 20 2020
(Python)
x, y, k, r, m = (3**n-3)//2, (3**n-3)//2, (n-1)%3, 3**(n-1), 0
while True:
m += 1
a, b = divmod(x, 3)
x, k = a+k*r, (k+k-b)%3
if y == x:
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CROSSREFS
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Cf. A001175 (period of Fibonacci numbers mod n).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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