A306352 a(n) is the least k >= 0 such that all the positive divisors of n have a distinct value under the mapping d -> d AND k (where AND denotes the bitwise AND operator).

0, 1, 2, 3, 4, 7, 2, 7, 10, 13, 2, 15, 4, 5, 6, 15, 16, 31, 2, 29, 6, 7, 2, 31, 12, 9, 10, 11, 4, 15, 2, 31, 42, 49, 6, 63, 4, 7, 6, 63, 8, 15, 2, 14, 14, 5, 2, 63, 18, 29, 18, 21, 4, 31, 6, 23, 18, 9, 2, 31, 4, 5, 14, 63, 76, 127, 2, 115, 6, 15, 2, 127, 8, 13

Offset: 1

Author

Rémy Sigrist, Feb 09 2019

Keywords

• nonn
• base

Comments

This sequence has similarities with A167234.

Will every nonnegative integer appear in the sequence?

Examples

			For n = 15:
- the divisors of 15 are: 1, 3, 5 and 15,
- their values under the mapping d -> d AND k for k = 0..6 are:
  k\d|  1  3  5  15
  ---+-------------
    0|  0  0  0  0
    1|  1  1  1  1
    2|  0  2  0  2
    3|  1  3  1  3
    4|  0  0  4  4
    5|  1  1  5  5
    6|  0  2  4  6
- the first row with 4 distinct values corresponds to k = 6,
- hence a(15) = 6.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Rémy Sigrist, Colored logarithmic scatterplot of the sequence for n = 1..2^19 (where the color is function of floor(n / 2^A070939(a(n)))).

Crossrefs

Cf. A000120, A002145, A028399, A070939, A131130, A167234, A218388.

Programs

PARI

a(n) = my (d=divisors(n)); for (m=0, oo, if (#Set(apply(v -> bitand(v, m), d))==#d, return (m)))

Formula

a(2^k) = 2^k - 1 for any k >= 0.

a(n) = 2 iff n belongs to A002145.

a(n) <= A218388(n).

a(n) AND A218388(n) = a(n).

A000120(a(n)) = 1 iff n is a prime number.

Apparently:

- a(3^k) belongs to A131130 for any k > 0,

- a(5^k) belongs to A028399 for any k >= 0.