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 A290772 Number of cyclic Gray codes of length 2n which include all-0 bit sequence and use the least possible number of bits. 0
 1, 2, 24, 12, 2640, 7536, 9408, 2688, 208445760, 1082368560, 4312566720, 12473296800, 24050669760 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS From Andrey Zabolotskiy, Aug 23 2017: (Start) The smallest number of bits needed is ceiling(log_2(n)). For larger number of bits, more Gray codes exist. Cyclic Gray codes of odd lengths do not exist, hence only even lengths are considered. A003042 is a subsequence: A003042(n+1) = a(2^n). a(n) is also the number of self-avoiding directed cycles of length 2n on a cube of the least possible dimension starting from the origin. (End) LINKS EXAMPLE Let n=3, so we count codes of length 6. Then at least 3 bits are needed to have such a code. There are a(3)=24 3-bit cyclic Gray codes of length 6: 000, 001, 011, 010, 110, 100 000, 001, 011, 111, 110, 100 000, 001, 011, 111, 110, 010 000, 001, 011, 111, 101, 100 000, 001, 101, 100, 110, 010 000, 001, 101, 111, 110, 100 000, 001, 101, 111, 110, 010 000, 001, 101, 111, 011, 010 000, 010, 011, 001, 101, 100 000, 010, 011, 111, 110, 100 000, 010, 011, 111, 101, 100 000, 010, 011, 111, 101, 001 000, 010, 110, 111, 101, 100 000, 010, 110, 111, 101, 001 000, 010, 110, 111, 011, 001 000, 010, 110, 100, 101, 001 000, 100, 101, 111, 110, 010 000, 100, 101, 111, 011, 010 000, 100, 101, 111, 011, 001 000, 100, 101, 001, 011, 010 000, 100, 110, 111, 101, 001 000, 100, 110, 111, 011, 010 000, 100, 110, 111, 011, 001 000, 100, 110, 010, 011, 001 PROG (Python) from math import log2, ceil def cyclic_gray(nb, n, a):     if len(a) == n:         if bin(a[-1]).count('1') == 1:             return 1         return 0     r = 0     for i in range(nb):         x = a[-1] ^ (1<

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Last modified January 18 13:09 EST 2019. Contains 319271 sequences. (Running on oeis4.)