推 cuttlefish :原來是取這樣的A_n跟M 學到了 10/10 12:30
: 00 P1=1
: 055 P2=3
: 2684 P3=4
: 134718976392 P4=12
: 02246066280886404482 P5=20
: 112358314594370774156178538190998752796516730336954932572910 P6=60
: 六個循環而已
: P1+...+P6=100
: ps 感覺跟群有關= = 但我代數不熟
A typical way to deal with Fibonacci-type numbers is
letting A_n=[a_n a_{n+1}] and M=[0 1]
[a_{n+1} a_{n+2}] [1 1]
then A_1*M^n = A_{n+1}
It's easy to see that M^3 = I mod 2, and M^20 = I mod 5.
Hence M^60 = I mod 10 ==> A_k=A_k*M^60 = A_{k+60}
Therefore, {a_i mod 10}_i is cyclic and its period is a factor of 60.
PS. If gcd(det(A_1),10)=1, then the period is 60.
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