then π(n) = 8k + 4. That is, if the modulo base is a Fibonacci number (≥ 3) with an even index, the period is twice the index and the cycle has two zeros. If the base is a Fibonacci number (≥ 5) with an odd index, the period is four times the index and the cycle has four zeros. Pisano periods of Lucas numbers If n = L(2k) (k ≥ 1), then π(n) = 8k; if n = L(2k + 1) (k ≥ 1), then π(n) = 4k + 2. That is, if the modulo base is a Lucas number (≥ 3) with an even index, the period is four times the index. If the base is a Lucas number (≥ 4) with an odd index, the period is twice the index. For even k, the cycle has two zeros. For odd k, the cycle has only one zero, and the second half of the cycle, which is of course equal to the part on the left of 0, consists of alternatingly numbers F(2m + 1) and n − F(2m), with m decreasing. Number of zeros in the cycle The number of occurrences of 0 per cycle is 1, 2, or 4. Let p be the number after the first 0 after the combination 0, 1. Let the distance between the 0s be q. There is one 0 in a cycle, obviously, if p = 1. This is only possible if q is even or n is 1 or 2. Otherwise there are two 0s in a cycle if p2 ≡ 1. This is only possible if q is even. Otherwise there are four 0s in a cycle. This is the case if q is odd and n is not 1 or 2. For generalized Fibonacci sequences (satisfying the same recurrence relation, but with other initial values, e.g. the Lucas numbers) the number of occurrences of 0 per cycle is 0, 1, 2, or 4. The ratio of the Pisano period of n and the number of zeros modulo n in the cycle gives the rank of apparition or Fibonacci entry point of n. That is, smallest index k such that n divides F(k). They are: 1, 3, 4, 6, 5, 12, 8, 6, 12, 15, 10, 12, 7, 24, 20, 12, 9, 12, 18, 30, 8, 30, 24, 12, 25, 21, 36, 24, 14, 60, 30, 24, 20,