Density of extended Mersenne numbers? Consider the subset of odd positive integers defined and constructed as follows by these rules :
A) $1$ is in the set.
B) if $x$ is in the set , then $2x + 1$ is in the set.
C) if $x$ and $y$ are in the set then $xy$ is in the set.
I call them extended Mersenne numbers because rule A and B alone give the Mersenne numbers $2^n - 1$.
And every product of Mersenne numbers must be in the set as well.
So the set or list of extended Mersenne numbers starts like
$$1,3,7,9,15,19,21,27,31,39,43,45,49,55,57,63,79,81,87,91,93,99,...$$
The main question is how dense is this within the odd integers ?
Secondary is there an easy and efficient test to know if an odd integer is in the list ?
I assume this sequence has no name yet but I could be wrong ?

*** speculation ***
I see $22$ odd elements from $1$ till $99$, which is close to the number of odd primes under $99$ (that is $24$)
$$ \dfrac11 + \dfrac13 + \dfrac17 + \dfrac19 + \ldots + \ldots+\dfrac{1}{99} = 2.034980..$$
So this (subjectively I admit ) seems to imply the sequence grows quite fast.
Therefore I guess maybe an asymptotic of the form
$C* x * \ln(x)^D $
for some constants $C,D$ might be an asymptotic for the counting function.
On the other hand I suspect much sharper asymptotics can be proven.


Background
This comes from abstract algebra where we want a commutative latin square with units and inverses such that "Property A"
$$ x * (y * y^{-1}) = x = (x * y) * y^{-1} $$
holds for all elements $x,y$.
( if this property has a name please tell me )
and for every element $z$
$$ z^2 = 1 $$
( not the main question here, but I wonder how dropping the $z^2 = 1$ condition effects things )
The dimensions of these are then exactly these extended Mersenne numbers + $1$.

edit
corrected forgotten terms and copy lulu's conjecture :
The sequence is equal to https://oeis.org/A197625 ???
( https://math.stackexchange.com/users/252071/lulu )
edit2
Greg Martin found counterexample 219 so lulu's conjecture is false.
Greg also conjectured that $D=0$ or $D$ goes to $0$.
It seems that Greg believes we will get close to linear, but I want to point out that the set is denser than the products of mersenne numbers.
On the other hand numbers like $63 = 3 * 21 = 2*31 + 1$ can be reached in $2$ ways so the rules have overlapping values, which could lower the density.
I would not be surprised if the density is slightly higher than the sum of 2 squares or the density of $a^3 + b^3 + c^3$ which are both less than linear.


It reminds me a bit of collatz, where the rules $2x$ and $(x-1)/3$ generate all integers , or so we believe.
But these rules for the extented Mersenne numbers have no deminishing rule like  $(x-1)/3$ ( deminish because that is smaller than $x$) , which is why I believe they might not have linear density.

added
Noticed how powers of $5$ or powers of $17$ are never in the list :
$5$ and $25$ are not in.
So $125,625,…$ are not created By products.
Also $125,625,…$ are of the form $4n + 1$, so they did not come from the 2x + 1 rule either.
Similar with $17$ or other primes missing of the form $4n + 1$.
This ofcourse highly influences the density.
Also notice primes of the form $4n+1$ have to come from 2x + 1 themselves , where x must be odd !!
2 x + 1 maps 1 mod 4 to 3 mod 4, and maps 3 mod 4 to 3 mod 4 !!
So primes 1 mod 4 are a key thing here !!

 A: Not an answer, but since the graphic may not fit a comment:
The below is a plot of the value $M(n)/n$, where $M(n)$ is the $n$-th value of the sequence (in ascending order, computed such that it's guaranteed to be complete up to $n=10^5$), for $n$ up to $10^5$. These ratios initially tend to rise, seemingly indicating that the set might not quite be of positive (lower) density, but surprisingly, from some point onwards they tend to come down again, and in particular, more than every seventh natural number seems to be in the set.

Update: Below, once again the same image, but this time up to $n=10^6$.
To give at least some shaky heuristic explanation for this behavior: the "average" number of divisors of an integer $n$, and with it the number of possible factorizations $n=ab$ into two factors is something like $\log(n)$. Hence if we assume that among integers smaller than $n$, the lower density of the sequence is some $c>0$, then with some probability $p(a)\cdot p(b)\approx c^2$, both of the given divisors $a,b$ are in the sequence (and hence so is $n$). But going over all divisors $a$, the probability that one never finds a pair $(a,b)$ both lying in the sequence becomes $\prod_{a|n} (1-p(a)\cdot p(n/a)) \approx (1-c^2)^{\log(n)}\to 0$. Of course this is completely unsound, since it assumes that the probabilities for the different divisors are independent, which they obviously aren't. But maybe it's possible to turn this into something sound (and indeed, in the above form, this would seem to indicate that the asymptotic proportion among odd integers is actually $1$, contrary to the first intuition, although this convergence would kick in extremely slow).

Update 2:
Some further plausibility testing, checking the probability for a number $x\in \{10^k+1,\dots, 10^k+10^5\}$ to belong to the sequence (guided by the hope that an interval of this length is somewhat representative for the probability of a number of that magnitude). For $k=6,7,\dots, 14$, I get the (rounded) values $6.536$,
$6.326$,
$6.138$,
$5.602$,
$5.397$,
$5.199$,
$5.038$,
$4.812$, $4.639$ respectively for the inverses of these probabilities, and it seems at least not completely implausible that these values might eventually converge to the (lowest possible) value $2$.
