

A329245


For any n > 0, let m = 2*n  1 (m is the nth 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
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OFFSET

1,1


COMMENTS

The sequence is well defined: for any n > 0:
 let x be such that 2*n1 < 2^x,
 hence gcd(2*n1, 2^x) = 1,
 and a(n) <= 1 + ord_{2^x}(2*n1) (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*n1, mk=m); for (k=2, oo, if (bitand(m, mk*=m)==m, return (k)))


CROSSREFS

Cf. A253719.
Sequence in context: A257246 A056206 A257245 * A155047 A029088 A253591
Adjacent sequences: A329242 A329243 A329244 * A329246 A329247 A329248


KEYWORD

nonn,base


AUTHOR

Rémy Sigrist, Nov 09 2019


STATUS

approved



