OFFSET
1,2
COMMENTS
Numbers n such that their binary weight is equal to their deficiency.
Numbers n such that A000120(n) = A033879(n), or equally A000203(n) = A005187(n), or equally A001065(n) = A011371(n).
2^(2^k-1) * (2^(2^k) + 1) is in the sequence if and only if (2^(2^k) + 1) is a (Fermat) prime (A019434) which as of today is known to be the case for 0 <= k <= 4 giving the terms 3, 10, 136, 32896 and 2147516416. - David A. Corneth, Nov 26 2017
It would be nice to know whether 315 is the only term that is neither in A191363 nor a power of two.
Any term that is either a square or twice a square (in A028982) must be odious (in A000069), and vice versa.
If there's an odd term below 10^30 besides 315 then it must be divisible by a prime >= 23. - David A. Corneth, Nov 27 2017
221753180448460815 is odd and also a term of this sequence. - Alexander Violette, Dec 24 2020
LINKS
EXAMPLE
A000203(315) = 1 + 3 + 5 + 7 + 9 + 15 + 21 + 35 + 45 + 63 + 105 + 315 = 624. 2*315 - 624 = 6, and when 315 is written in binary, 100111011, we see that it has six 1-bits. Thus 315 is included in the sequence.
MATHEMATICA
Select[Range[2^20], DivisorSigma[1, #] + DigitCount[#, 2, 1] == 2 # &] (* Michael De Vlieger, Nov 26 2017 *)
PROG
(PARI) for(n=1, oo, if(sigma(n)+hammingweight(n) == 2*n, print1(n, ", ")));
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Antti Karttunen, Nov 26 2017
EXTENSIONS
Terms a(35) and beyond from Giovanni Resta, Feb 27 2020
STATUS
approved