A337212 Modulo 3 Pisano period of 'n-bonacci' series.

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

Offset: 1

Author

Adam Bascal, Aug 19 2020

Keywords

• nonn

Comments

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

Examples

			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.
		

Crossrefs

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

Programs

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

def A337212(n):
    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:
            return m # Chai Wah Wu, Sep 14 2020

Formula

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

Extensions

a(20)-a(22) from Jinyuan Wang, Aug 20 2020

a(23) from Chai Wah Wu, Sep 14 2020

a(24)-a(28) from Chai Wah Wu, Sep 15 2020

a(29)-a(31) from Chai Wah Wu, Sep 21 2020