A294523 Lexicographically earliest sequence of positive terms, such that, for any n > 0, the binary expansion of n, say of size k+1, is (1, a(n) mod 2, a^2(n) mod 2, ..., a^k(n) mod 2) (where a^i denotes the i-th iterate of the sequence).

1, 2, 1, 2, 6, 5, 1, 2, 10, 6, 14, 9, 5, 13, 1, 2, 18, 10, 22, 12, 6, 14, 30, 17, 9, 5, 11, 25, 13, 29, 1, 2, 34, 18, 38, 20, 10, 22, 46, 24, 12, 6, 54, 28, 14, 30, 62, 33, 17, 9, 19, 41, 5, 11, 23, 49, 25, 13, 27, 57, 29, 61, 1, 2, 66, 34, 70, 36, 18, 38, 78

Offset: 1

Author

Rémy Sigrist, Nov 01 2017

Keywords

• nonn
• base
• look

Comments

More informally, the parity of the iterate of the sequence at n gives the binary expansion of n (beyond the leading 1).

Apparently, iterating the sequence always leads to one of these three loops:

- the fixed point (1) iff we start from 2^k-1 for some k > 0,

- the fixed point (2) iff we start from 2^k for some k > 0,

- or (5, 6) for any other starting value.

a(n) is even iff n belongs to A004754.

a(n) is odd iff n belongs to A004760.

If a(n) > n then a(n) = A080541(n).

If n < 2^k then a(n) < 2^k.

Apparently, if a(n) > 2, then A054429(a(n)) = a(A054429(n)); this accounts for the symmetry of the part connected to the loop (5,6) in the oriented graph of this sequence.

Examples

			For n=11:
- the binary representation of 11 is (1,0,1,1),
- a(11) = 14 has parity 0,
- a(14) = 13 has parity 1,
- a(13) = 5 has parity 1,
- we find the binary digits of 11 beyond the initial 1, in order: 0, 1, 1.
See also representations of first terms in Links section.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..8192
• Rémy Sigrist, Representation of the first 32 terms as an oriented graph (n -> a(n))
• Rémy Sigrist, Representation of the first 64 terms as an oriented graph (n -> a(n))
• Rémy Sigrist, PARI program for A294523

Crossrefs

Cf. A000079, A000225, A000975, A004754, A004760, A052997, A054429, A080541, A081253, A081254, A266248, A266613, A266721, A267045.

Programs

PARI

See Links section.

Formula

a(n) = 1 iff n = A000225(k) for some k > 0.

a(n) = 2 iff n = A000079(k) for some k > 0.

a(n) = 5 iff n = A081254(k) for some k > 2.

a(n) = 6 iff n = A000975(k) for some k > 2.

a(n) = 10 iff n = A081253(k) for some k > 2.

a(n) = 12 iff n = A266613(k) for some k > 3.

a(n) = 13 iff n = A052997(k) for some k > 2.

a(n) = 14 iff n = A266721(k) for some k > 2.

a(n) = 18 iff n = A267045(k) for some k > 3.

a(n) = 54 iff n = A266248(k) for some k > 4.

These formulas come from the fact that each sequence on the right side, say f, eventually satisfies: f(n) = floor(f(n+1)/2), and f(n) and f(n+2) have the same parity.