A266351 Start with a(1) = 1, then always choose for a(n) the least unused number such that A057889(a(n)*a(n-1)) = A057889(a(n)) * A057889(a(n-1)), where A057889 is a bijective base-2 reverse.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 11, 17, 13, 32, 15, 24, 18, 20, 19, 33, 21, 36, 28, 27, 56, 34, 22, 64, 23, 65, 25, 40, 35, 72, 42, 48, 30, 51, 60, 66, 26, 68, 37, 96, 31, 99, 62, 128, 29, 129, 38, 80, 49, 73, 70, 130, 39, 256, 41, 131, 74, 136, 44, 132, 46, 257, 43, 258, 45, 260, 47, 512, 50, 133, 76, 160, 67, 84, 97, 137, 112, 54

Offset: 1

Author

Antti Karttunen, Dec 28 2015

Keywords

• nonn
• base

Comments

Equally: always choose for a(n) the least unused number such that a(n)*a(n-1) = A057889(A057889(a(n)) * A057889(a(n-1))).

Note that the adjacent terms of permutation A266195 satisfy the same condition, except that permutation is not the lexicographically earliest sequence of this kind (because it has a more restrictive condition). See A266194.

This is a bijection for the same reason that A266195 is. Any high enough 2^k will always save the permutation of being stuck, and will also immediately pick up as its succeeding pair the least term unused so far.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..1450
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Inverse: A266352.

Cf. A057889, A266194.

Cf. A266195, A265405, A266405 (similar sequences).