A359508 a(n) = log_2(A359507(n) - 1).

0, 1, 1, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2

Offset: 1

Author

Peter Kagey, Jan 03 2023

Keywords

• nonn

Comments

Conjecture: A359507(n) is always of the form 2^m + 1.

If log_2(A359507(n) - 1) is not an integer, then define a(n) = -1.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A000523, A359506, A359507, A378218.

Programs

PARI

A359506(n) = if(n==0, return (0), my (x=[n], y); for (m=n+1, oo, if (vecmin(y=[bitxor(v, m) | v<-x])==0, return (m), x=setunion(x, Set(y))))); \\ From A359506.
A359507(n) = (A359506(n)-n);
A359508(n) = (#digits(A359507(n)-1, 2)-1); \\ Antti Karttunen, Nov 22 2024

Formula

a(n) = A000523(A359507(n)-1).

Conjecture:

a(4k) = 1 for k > 0,

a(4k+1) = 2 for k > 0,

a(4k+2) = 1 for k > 0,

a(4k+3) = a(k) + 2 for k > 0.

Apparently, a(n) = abs(A378218(1+n)). [This holds at least up to n=65537] - Antti Karttunen, Nov 22 2024

a(n) = A007814((n - 3*b(n + 1) + 2) mod b(n + 1) + b(n + 2) - 1) + 1, where b(n) = 2^A000523(A002264(n)) for n >= 4. - Alan Michael Gómez Calderón, Feb 25 2025

Extensions

More terms from Antti Karttunen, Nov 22 2024