Through mathematical induction, we should know a lemma, which is well proved by induction: when n≥2, * * has n (n+ 1) (n- 1)/6 lengths of n and is exactly1"01". When the length is n 4, two "0 1" are determined first, and the remaining length is n, then these n bits either have no "0 1" (at this time, these two "0 1" are arranged with n individuals as two identical points; There are 0000, 1000, 1 100,110, 165438+ with n 2 positions. Choose two and put two "0 1", and the rest are dead, so * * has (n 1)C(n 2, 2)) or a "0 1" (then this "0 1" is broken. Another is to put a "0 1" into that "0 1", take the remaining "0 1" as an individual, and take that "00 1 1" as an individual, leaving n-2 individuals, n *.