Subpolynomial growth of the last near-perfect-power index

Prove that for every integer k > 1, the quantity M_k(d), defined as the largest n for which A_k(n) <= d, satisfies M_k(0) = 1 and M_k(d) = o(d^epsilon) for every epsilon > 0.

Background

Assuming the finiteness conjecture, the paper introduces M_k(d) as the last index n for which p(n) lies within distance d of a kth power. The authors use numerical data to propose a refined description of its growth.

The first assertion recovers Sun’s conjecture at distance zero, while the second asserts that M_k(d) grows more slowly than every positive power of d.

References

Based on these numerics, we make the following refinement of Conjecture 1. Conjecture 2. If k > 1, then the following are true. (1) We have that M_k(0) = 1. (2) For every ε > 0, we have that M_k(d) = o(dε).

— Do perfect powers repel partition numbers?  (2501.03754 - Merca et al., 7 Jan 2025) in Section 2, “Further Conjectures: Do kth Powers Repel Partition Numbers?”, Conjecture 2