A302439 a(n) is the number of ways of writing the binary expansion of n as a concatenation of nonempty aperiodic substrings (i.e., substrings that are not the concatenation of 2 or more equal parts).

1, 1, 2, 1, 3, 3, 3, 1, 4, 4, 6, 4, 5, 5, 4, 1, 5, 5, 10, 5, 10, 9, 11, 5, 7, 7, 10, 7, 7, 7, 5, 1, 6, 6, 14, 6, 16, 15, 16, 6, 14, 14, 19, 13, 18, 17, 16, 6, 9, 9, 17, 9, 16, 15, 17, 9, 10, 10, 14, 10, 9, 9, 6, 1, 7, 7, 18, 7, 24, 21, 21, 7, 23, 22, 32, 22

Offset: 0

Author

Rémy Sigrist, Apr 08 2018

Keywords

• nonn
• base

Comments

Leading zeros in the binary expansion of n are ignored.

The value a(0) = 1 corresponds to the empty concatenation.

See A301453 for similar sequences.

For some odd o = 2*k + 1, is there some k such that for all e > k, a(o * 2^e) = a(o + 2^(e - 1)) + c for some c? - David A. Corneth, Apr 15 2018

Examples

			For n = 20: the binary expansion of 20, "10100", can be split in 10 ways into aperiodic substrings:
- (10100),
- (101)(0)(0),
- (10)(100),
- (10)(10)(0),
- (10)(1)(0)(0),
- (1)(0100),
- (1)(010)(0),
- (1)(0)(100),
- (1)(0)(10)(0),
- (1)(0)(1)(0)(0).
Hence a(20) = 10.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..10000
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A301453.

Programs

PARI

a(n) = if (n==0, return (1), my (v=0); for (w=1, #binary(n), my (ok=1); fordiv (w, d, if (d
				

Formula

a(2^n - 1) = 1 for any n >= 0.

a(2^n) = n + 1 for any n >= 0.

From David A. Corneth, Apr 15 2018: (Start)

Is a(2^n + i) >= a(2^n) for 0 <= i <= 2^n - 2?

What is the least k(n) such that

a(2^n + i) <= a(2^n + k(n)) for 1 <= i <= 2^n - 2? (End)