Does the reciprocal sum of primitive pseudoperfect numbers converge? Compute the partial sums (Erdős #469)

Call $n$ pseudoperfect (semiperfect) if $n=d_1+\cdots+d_k$ for some distinct proper divisors $d_i$ of $n$. Let $A$ be the set of PRIMITIVE pseudoperfect numbers: those $n$ that are pseudoperfect but for which no proper divisor $m\mid n$ with $m<n$ is itself pseudoperfect. Erdős asked whether $\sum_{n\in A}\frac1n$ converges. GOAL (computational evidence): enumerate $A\cap[1,N]$ for $N$ as large as feasible, compute the partial sums $\sum_{n\in A, n\le N}\frac1n$ and the counting function $|A\cap[1,N]|$, and report their growth — a slowly-growing-but-unbounded vs. plateauing partial sum is direct evidence on convergence. Testing whether a given $n$ is pseudoperfect is a subset-sum over its divisors (bounded, decidable).

Acceptance. PARTIAL / PRIMARY (finite, machine-checkable): a correct enumeration of the primitive pseudoperfect numbers up to $N$ (state $N$; extending OEIS A006036 beyond its current range is itself a result), delivered with (i) a runnable subset-sum-based membership test, (ii) the partial reciprocal sum $\sum_{n\in A,n\le N}1/n$ to stated precision, and (iii) a fit of the counting function $|A\cap[1,N]|$. Cross-check against known A006036 terms on the overlap. FULLY RESOLVES (analytic): a proof that $\sum_{n\in A}1/n$ converges or diverges. State $N$ and the divisor/subset-sum method.