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

1,1

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).

Links

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

Example

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.

Prog

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

Crossrefs

Cf. A253719.

Sequence in context: A257246 A056206 A257245 * A155047 A369451 A029088

Adjacent sequences: A329242 A329243 A329244 * A329246 A329247 A329248

Keyword

nonn,base

Author

Rémy Sigrist, Nov 09 2019

Status

approved