A253719 -id:A253719 - cp's OEIS frontend

A329245 For any n > 0, let m = 2*n - 1 (m is the n-th odd number); a(n) is the least k > 1 such that m AND (m^k) = m (where AND denotes the bitwise AND operator).

2, 3, 3, 3, 3, 5, 5, 3, 3, 7, 7, 5, 5, 9, 9, 3, 3, 3, 11, 3, 3, 5, 5, 5, 5, 15, 7, 9, 9, 17, 17, 3, 3, 3, 5, 5, 5, 5, 9, 3, 3, 23, 7, 13, 13, 9, 9, 5, 5, 19, 11, 3, 3, 5, 21, 9, 9, 15, 23, 17, 17, 33, 33, 3, 3, 3, 3, 3, 7, 5, 5, 3, 3, 7, 7, 21, 21, 17, 9, 3, 3

Offset: 1

Graph
List
Refs
Listen
History
Text
Internal format

Rémy Sigrist, Nov 09 2019

nonn
base

The sequence is well defined: for any n > 0:
- let x be such that 2*n-1 < 2^x,
- hence gcd(2*n-1, 2^x) = 1,
- and a(n) <= 1 + ord_{2^x}(2*n-1) (where ord_u(v) is the multiplicative order of v modulo u).

			For n = 7:
- m = 2*7 - 1 = 13,
- 13 AND (13^2) = 9,
- 13 AND (13^3) = 5,
- 13 AND (13^4) = 1,
- 13 AND (13^5) = 13,
- hence a(7) = 5.

Rémy Sigrist, Table of n, a(n) for n = 1..8192

Cf. A253719.

a(n) = my (m=2*n-1, mk=m); for (k=2, oo, if (bitand(m, mk*=m)==m, return (k)))

cp's OEIS Frontend

A329245 For any n > 0, let m = 2*n - 1 (m is the n-th odd number); a(n) is the least k > 1 such that m AND (m^k) = m (where AND denotes the bitwise AND operator).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI