A093696 Numbers n such that all divisors of n have an odd number of 1's in their binary expansions.
1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 22, 26, 28, 31, 32, 37, 38, 41, 44, 47, 49, 52, 56, 59, 61, 62, 64, 67, 73, 74, 76, 79, 82, 88, 91, 94, 97, 98, 103, 104, 107, 109, 112, 118, 121, 122, 124, 127, 128, 131, 133, 134, 137, 143, 146, 148, 151, 152, 157, 158, 164, 167, 173
Offset: 1
Examples
14 is in the sequence because its divisors are [1, 2, 7, 14] and in binary: 1, 10, 111 and 1110, all have an odd number of 1's.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Maple
isA001969 := proc(n) if wt(n) mod 2 = 0 then true; else false; end if; end proc: isA093696 := proc(n) for d in numtheory[divisors](n) do if isA001969(d) then return false; end if; end do; true; end proc: for n from 1 to 200 do if isA093696(n) then printf("%d,",n); end if; end do: # R. J. Mathar, Feb 13 2014
-
Mathematica
odiousQ[n_] := OddQ @ DigitCount[n, 2][[1]]; Select[Range[200], AllTrue[ Divisors[#], odiousQ ] &] (* Amiram Eldar, Dec 09 2019 *)
-
PARI
is(n)=fordiv(n,d,if(hammingweight(d)%2==0, return(0))); 1 \\ Charles R Greathouse IV, Mar 29 2013
-
Python
from sympy import divisors, isprime def c(n): return bin(n).count("1")&1 def ok(n): return n > 0 and all(c(d) for d in divisors(n, generator=True)) print([k for k in range(174) if ok(k)]) # Michael S. Branicky, Jul 24 2022
Formula
{n: A356018(n) =0 }. - R. J. Mathar, Aug 07 2022
Comments