Erdős Problem 1134: a collision gives density zero
Statement and scope.
Let $A$ be the least set of positive integers containing $1$ and closed under $x\mapsto2x+1,3x+1,6x+1$. Does $A$ have positive lower density? No: we prove the quantitative bound $A(X)=O(X^{99/100})$. This is a reconstruction of the known negative resolution, not a claim to improve the published exponent. See
https://www.erdosproblems.com/1134.
A shortening identity.
Write a word in the alphabet $\{2,3,6\}$ in the order in which its maps $F_m(x)=mx+1$ are applied. Then \[
F_2(F_2(F_3(x)))=12x+7=F_6(F_2(x)).
\] Consequently the consecutive subword $322$ may be replaced by $26$ without changing the represented integer or the product of its letters. Every value in $A$ therefore has a representing word with no occurrence of $322$: repeated replacements terminate because they shorten the word. The empty word represents $1$.
A convergent word sum.
Put $\sigma=99/100$ and $S=2^{-\sigma}+3^{-\sigma}+6^{-\sigma}$. A word avoiding $322$ avoids it in particular in each of its disjoint blocks of three letters. If $P(w)$ is the product of the letters, this gives \[
\sum_{\substack{w:\,322\text{ absent}}}P(w)^{-\sigma}
\leq (1+S+S^2)\sum_{j\geq0}(S^3-12^{-\sigma})^j.
\] The right side is finite. Indeed \[
S\leq6^{1/100}(1/2+1/3+1/6)=6^{1/100},
\qquad 6^{3/100}<13/12,\qquad 12^{-\sigma}>1/12.
\] The middle strict inequality is equivalently the exact integer inequality $216\cdot12^{100}<13^{100}$. Thus $S^3-12^{-\sigma}<1$.
Counting distinct integers.
Starting from $1$, each operation increases the value at least by its multiplicative factor, so the value represented by $w$ is at least $P(w)$. Choose one shortened word for each member of $A\cap[1,X]$. These words are distinct and satisfy $P(w)\leq X$. Therefore, writing $C$ for the finite word sum above, \[
A(X)\leq\sum_{\substack{w:\,322\text{ absent}\\P(w)\leq X}}1
\leq X^\sigma\sum_{w:\,322\text{ absent}}P(w)^{-\sigma}
\leq CX^\sigma.
\] In particular the natural density exists and is zero.
Attribution and dependencies.
The problem page reports the Crampin--Hilton resolution and a sharper exponent, also discussed in Lagarias's account. This proof uses the same basic phenomenon of coincident affine words, with an elementary block estimate; it needs no external counting theorem.
Completion estimate.
The requested positive-density assertion is completely disproved.
COMPLETION: 100%