A074203 Odd numbers k such that the number of 1's in the binary representation of k divides 2^k-1.

1, 351, 375, 381, 471, 477, 501, 687, 699, 747, 855, 861, 885, 939, 981, 1119, 1143, 1149, 1239, 1245, 1269, 1311, 1335, 1341, 1359, 1371, 1383, 1389, 1395, 1401, 1431, 1437, 1461, 1479, 1485, 1491, 1497, 1509, 1521, 1623, 1629, 1653, 1707, 1749, 1815

Offset: 1

Author

Benoit Cloitre, Sep 17 2002

Keywords

• base
• easy
• nonn

Comments

Except for 1, terms seem always divisible by 3.

From Robert Israel, Jan 14 2019: (Start)

An odd number k is in the sequence if and only if A000120(k) is in A036259 and k is divisible by A007733(A000120(k)). In particular, there are infinitely many of these for every member of A036259 except 1.

Thus a(2) to a(28842) have A000120(k)=7 and are divisible by 3, but a(28843) = 12582911 has A000120(12582911) = 23 and is divisible by A007733(23) = 11 but not by 3. (End)

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000120, A007733, A036259, A074202.

Programs

Maple

filter:= n -> 2 &^ n - 1 mod convert(convert(n,base,2),`+`) = 0:
select(filter, [seq(i,i=1..2000,2)]); # Robert Israel, Jan 13 2019

Mathematica

Join[{1}, Select[Range[3, 2000, 2], PowerMod[2, #, DigitCount[#, 2, 1]] == 1 &]] (* Amiram Eldar, Jun 08 2022 *)

PARI

isok(n) = (n % 2) && !((2^n-1) % hammingweight(n)); \\ Michel Marcus, Nov 29 2013