A295296 Numbers n such that the sum of their divisors + the number of ones in their binary expansion = 2n; numbers for which A000203(n) + A000120(n) = 2n.

1, 2, 3, 4, 8, 10, 16, 32, 64, 128, 136, 256, 315, 512, 1024, 2048, 4096, 8192, 16384, 32768, 32896, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 2147516416

Offset: 1

Author

Antti Karttunen, Nov 26 2017

Keywords

• nonn
• base

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

Examples

			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.
		

Links

• Max Alekseyev, Table of n, a(n) for n = 1..70 (first 52 terms from Giovanni Resta)
• Index entries for sequences related to binary expansion of n
• Index entries for sequences related to sigma(n)

Crossrefs

Positions of zeros in A294898 and A294899.

Subsequence of A005100 and also of A295298.

Subsequences: A000079, A191363 (the five known terms).

Cf. A000120, A000203, A001065, A005187, A011371, A019434, A033879, A141548.

Programs

Mathematica

Select[Range[2^20], DivisorSigma[1, #] + DigitCount[#, 2, 1] == 2 # &] (* Michael De Vlieger, Nov 26 2017 *)

PARI

for(n=1, oo, if(sigma(n)+hammingweight(n) == 2*n, print1(n, ", ")));

Extensions

Terms a(35) and beyond from Giovanni Resta, Feb 27 2020