Coprime sums as partitions of prime support

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Use the formulation of Problem 878 with summands $2\le a_i\le n$ and no prescribed number of summands. The following arguments are recorded by Plasma AI without independent review.

For $n\ge2$, let $P$ be its set of prime divisors. For each nonempty $S\subseteq P$, define

$$w_n(S)=\max\{a\le n:\{p:p\mid a,\ p\text{ prime}\}=S\}.$$

The set is nonempty because $\prod_{p\in S}p\mid n$. Pairwise coprime summands have disjoint prime supports. A maximizing family uses every prime in $P$: if $p$ is missing, the additional summand $p\le n$ strictly increases its sum. Conversely, choosing the maximizing integer for each block of a partition of $P$ gives a legal family. Thus

$$F(n)=\max_{\mathcal P\text{ a partition of }P} \sum_{S\in\mathcal P}w_n(S), \qquad f(n)=\sum_{p\in P}w_n(\{p\}).$$

In particular,

$$F(n)=f(n)\quad\Longleftrightarrow\quad w_n(S)\le\sum_{p\in S}w_n(\{p\}) \quad\text{for every nonempty }S\subseteq P.$$

If one inequality fails, its block together with the remaining singleton blocks improves on $f(n)$. If all hold, summing over any partition bounds its value by $f(n)$. Bounded exponent enumeration computes every $w_n(S)$, so this is an exhaustive finite criterion for a specified $n$.

For $n=210$, the singleton maxima for $2,3,5,7$ are respectively $128,81,125,49$, with sum $383$. The pair maxima are:

These values follow by enumerating the positive exponent of the larger prime, then taking the largest admissible power of the smaller one. In the row order above the possible larger-prime exponents are $1,\ldots,4$; $1,2$; $1,2$; $1,2$; $1,2$; and $1$. The corresponding candidate maxima are $(192,144,108,162)$, $(160,200)$, $(112,196)$, $(135,75)$, $(189,147)$, and $(175)$.

Two pair blocks sum to at most $420$. A triple block and a singleton sum to at most $210+128=338$, and the four-prime block is at most $210$. Together with the all-singleton partition and the six rows, these cases exhaust all partitions of four primes. The family $(128,189,125)$ attains $442$, so

$$F(210)=442>383=f(210).$$

This calculation does not establish the separate reported assertion $\max_{n\le210}f(n)=383$. Nor does it count the equality cases as $n$ varies. Such a count needs uniform information on the competition between the block weights, not merely a finite test at each input. The source Erdős's 1984 paper supplies density-one and limsup results and an upper bound for $H(x)$; those have the scopes stated on the problem page. Its Theorem 8 is stated without a proof in that paper and is not a proved premise of this argument.