A337212 - OEIS

%I #47 Dec 24 2022 02:30:21

%S 1,8,13,26,104,728,364,80,91,8744,3851,3280,59048,4782968,7174453,

%T 3438578,16139240,5373368,5235412,1678822106,86049704,387420488,

%U 47071589413,140633637386,2952400,728,757,9526168288,7312949144072,49566102697280,24477226494760

%N Modulo 3 Pisano period of 'n-bonacci' series.

%C 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).

%C 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

%F 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

%e 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.

%o (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

%o (Python)

%o def A337212(n):

%o     x, y, k, r, m = (3**n-3)//2, (3**n-3)//2, (n-1)%3, 3**(n-1), 0

%o     while True:

%o         m += 1

%o         a, b = divmod(x,3)

%o         x, k = a+k*r, (k+k-b)%3

%o         if y == x:

%o             return m # _Chai Wah Wu_, Sep 14 2020

%Y Cf. A001175 (period of Fibonacci numbers mod n).

%K nonn

%O 1,2

%A _Adam Bascal_, Aug 19 2020

%E a(20)-a(22) from _Jinyuan Wang_, Aug 20 2020

%E a(23) from _Chai Wah Wu_, Sep 14 2020

%E a(24)-a(28) from _Chai Wah Wu_, Sep 15 2020

%E a(29)-a(31) from _Chai Wah Wu_, Sep 21 2020