OFFSET
1,2
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
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
EXAMPLE
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.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Adam Bascal, Aug 19 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
STATUS
approved