Inspired by this question , where it is asked for positive integers with the property $$2n-\sigma(n)\mid \sigma(n)-n$$ which is equivalent to $$2n-\sigma(n)\mid n$$ The author also demands $2n-\sigma(n)>1$. $\sigma(n)$ is the divisor-sum function.
Question : Let $S$ be the set of all positive integers with the above property. Is it true, that for every non-negative integer $k$ , there is a number $n=2^k\cdot m$ with some odd integer $m>1$ , such that $n\in S$ , in other words is there a solution whatever $k$ is ?
I used the following PARI/GP program to find the smallest solution, if it does not exceed $10^7$
check(n)={h=0;s=sigma(n);if(2*n>s+1,if(Mod(n,2*n-s)==0,h=1));h}
gp > for(k=0,23,m=3;while((m<10^7)*(check(2^k*m)==0),m=m+2);if(check(2^k*m)==1,print(k," ",m));if(check(2^k*m)==0,print(k," ?")))
0 9018009
1 5
2 11
3 17
4 47
5 67
6 131
7 257
8 1051159
9 1031
10 2063
11 4099
12 8447
13 16447
14 32771
15 65537
16 ?
17 262147
18 524351
19 1048583
20 2101247
21 4194319
22 8388623
23 ?
As one can see, for $k=16$ and for $k=23$ , there is no solution below $10^7$.
In the special case that $m$ is a prime number , we have $$\sigma(n)=(2^{k+1}-1)\cdot (m+1)$$ hence $$2n-\sigma(n)=m+1-2^{k+1}$$ If this is $2^s$ with some positive integer $s\le k$, then we have found a solution with $m:=2^{k+1}+2^s-1$
This gives solutions for example for $k=23$ and $k=25$ , but not for $k=16$ and $k=24$.
Update : The smallest cases without a known example are now $$k=164,172,192,200$$