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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).
1
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
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
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, Nov 09 2019
STATUS
approved