Reciprocal Mersenne Subseries

Theorem 6.108 (A restriction on dyadic rational values). If an infinite support $A\subseteq {\mathbb{N}}_{>0}$ has $X_A(2)=p/2^c$ for integers $p$ and $c\ge 0$, then $\sum _{a\in A}1/a$ either diverges or converges to a value greater than $1$. The cited proof averages the shifted integer recurrence and uses a common multiple of two distinct support elements. Thus a convergent reciprocal sum of at most $1$ is excluded. This is a separate necessary condition: the reciprocal-summable criterion proved earlier already excludes every support with a convergent reciprocal sum, not just those whose sum is at most $1$.

Theorem 6.109 (Unboundedness of a positive shifted recurrence). Let $A\subseteq {\mathbb{N}}_{>0}$ be infinite, fix a shift $c\ge 0$ and a positive integer $v$, and let $u$ be a positive integer sequence satisfying

Then $u$ is unbounded. Indeed, under a proposed bound $u(n)\le B$, choose $2B+1$ distinct support elements and a common multiple larger than $c$. At the corresponding index the divisor count is at least $2B+1$, contradicting the recurrence. This does not conflict with Theorem 6.105, whose condition is $u(n)/2^n\to 0$, not boundedness. An unbounded sequence can still satisfy that limit.

Theorem 6.110 (Bounds for intervals with zero divisor counts). Suppose that $X_A(2)=p/(2^{c}v)$ for an infinite positive support $A$ and an odd positive integer $v$. For every $\epsilon >0$ there is $B$ such that, for $N\ge 1$, a run of $h$ zero divisor counts starting after $c+N$ satisfies $h\le \epsilon {\log }_{2}N+B$. In fact this conclusion holds without rationality: fix any $a\in A$. Every $a$ consecutive positive integers include a multiple of $a$, where $c_A$ is positive. Hence $h\le a-1$, so one can take $B=a-1$. The zero-run conclusion therefore imposes no additional restriction on a fixed nonempty support, regardless of the tail estimates used in the linked proof.