A290772 Number of cyclic Gray codes of length 2n which include all-0 bit sequence and use the least possible number of bits.
1, 2, 24, 12, 2640, 7536, 9408, 2688, 208445760, 1082368560, 4312566720, 12473296800, 24050669760, 27034640640, 13900259520, 1813091520
Offset: 1
Examples
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
Links
- Thomas König, Fortran program for counting
Programs
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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<
Extensions
a(7)-a(8) and name from Andrey Zabolotskiy, Aug 23 2017
a(9)-a(13) from Ashis Kumar Mal, Sep 02 2017
a(14)-a(16) from Thomas König, Jan 22 2022
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