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

Author

Rémy Sigrist, Nov 09 2019

Keywords

• nonn
• base

Comments

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).

Examples

			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.
		

Links

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

Crossrefs

Cf. A253719.

Programs

PARI

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