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%I #27 Jun 08 2022 03:20:14
%S 1,351,375,381,471,477,501,687,699,747,855,861,885,939,981,1119,1143,
%T 1149,1239,1245,1269,1311,1335,1341,1359,1371,1383,1389,1395,1401,
%U 1431,1437,1461,1479,1485,1491,1497,1509,1521,1623,1629,1653,1707,1749,1815
%N Odd numbers k such that the number of 1's in the binary representation of k divides 2^k-1.
%C Except for 1, terms seem always divisible by 3.
%C From _Robert Israel_, Jan 14 2019: (Start)
%C 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.
%C 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)
%H Robert Israel, <a href="/A074203/b074203.txt">Table of n, a(n) for n = 1..10000</a>
%p filter:= n -> 2 &^ n - 1 mod convert(convert(n,base,2),`+`) = 0:
%p select(filter, [seq(i,i=1..2000,2)]); # _Robert Israel_, Jan 13 2019
%t Join[{1}, Select[Range[3, 2000, 2], PowerMod[2, #, DigitCount[#, 2, 1]] == 1 &]] (* _Amiram Eldar_, Jun 08 2022 *)
%o (PARI) isok(n) = (n % 2) && !((2^n-1) % hammingweight(n)); \\ _Michel Marcus_, Nov 29 2013
%Y Cf. A000120, A007733, A036259, A074202.
%K base,easy,nonn
%O 1,2
%A _Benoit Cloitre_, Sep 17 2002
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