Zhai 1999 additive completion kth powers

../

corollary_p293: Zhai's two-sided estimate for completions of the kth powers up to N confined to [0, delta N]: for N >= N_k and M_k/N < delta < delta_k, f_k(delta N, N) lies between (k - 3k^{3/2} delta^{1/2}) N^{1-1/k} and k(1 + k N^{-1/k}) N^{1-1/k}.

proposition_p292: Zhai's two-sided bound on the smallest M for which some subset of [0, M] completes the kth powers up to N, in terms of B, the integer part of the kth root of N.

theorem_1: Zhai's lower bound for completions of the kth powers confined to a short initial interval: for N at least N_0(k) and 3k^2 N^{-1/2k} <= eps <= eps_0(k), f_k(delta N, N) >= (k - eps) N^{1-1/k} with delta = eps^2/(9k^3).

theorem_2: Zhai's upper bound for the least size of a set in [0, M] completing the kth powers up to N, valid for every M from M_k(N) to N, with B the integer part of the kth root of N.

theorem_3: Zhai's lower bound for completions of the kth powers up to N confined to [0, delta N] with delta fixed in (0,1) and k large, with an explicit constant C(delta) and exponent Delta = log(1/delta)/log(1+1/delta).

---

Wenguang Zhai, The Additive Completion of kth Powers. Journal of Number Theory 79 (1999), 292-300. doi:10.1006/jnth.1999.2441.

For fixed k >= 2 let f_k(M,N) be the least size of a set A contained in [0,M] such that every positive integer n <= N can be written as a + b^k with a in A and b a positive integer. Earlier work treated M = N, the latest such bound being Cilleruelo's (1). Zhai instead localizes A to a short interval. Theorem 1 gives constants eps_0(k) and N_0(k) such that for N >= N_0 and 3k^2 N^{-1/2k} <= eps <= eps_0 one has f_k(delta N, N) >= (k-eps)N^{1-1/k} with delta = eps^2/(9k^3), and Theorem 2 supplies the matching upper bound f_k(M,N) <= (B+1)^k - B^k with B = floor(N^{1/k}) for every M from the least admissible value M_k(N) up to N; the Corollary combines them into (k - 3k^{3/2}delta^{1/2})N^{1-1/k} <= f_k(delta N,N) <= k(1+kN^{-1/k})N^{1-1/k} for N >= N_k and M_k(N)/N < delta < delta_k, hence the asymptotic f_k(delta N,N) = (k+o(1))N^{1-1/k} when delta = delta(N) tends to 0 as N tends to infinity. A Proposition bounds the least admissible M between B^k-(B-1)^k-1 and (B+1)^k-B^k-1, and Theorem 3 gives, for fixed delta in (0,1) and large k, a lower bound of order C(delta)(k^Delta/log k)N^{1-1/k}, where Delta = log(1/delta)/log(1+1/delta) lies strictly between 0 and 1. The author remarks that the asymptotic may be conjectured to hold uniformly for M_k(N) <= M <= N, while the method works only for M = o(N), and that the theorems also hold with b^k replaced by b^k + P_{k-1}(b), P_{k-1} a polynomial of degree k-1, or by [b^c] with c >= 2 a fixed real number (p. 294). For problem 33 this is a follow-up to Cilleruelo's bound, but because it concerns finite complements chosen separately for each N and localized to a short interval, it does not determine either counting constant that problem 33 asks about.

Source: https://doi.org/10.1006/jnth.1999.2441. The print carries "Copyright © 1999 by Academic Press" and "All rights of reproduction in any form reserved." in the footer of its first page (printed p. 292), every other right reserved.

Bears on. #33: at k = 2 the Corollary (p. 293) gives f_2(delta N, N) = (2+o(1))N^(1/2) when delta = delta(N) tends to 0 as N tends to infinity, f_2(delta N, N) being the least size of a set in [0, delta N] completing the squares b^2, b >= 1, up to N, and Theorem 2 (p. 293) gives f_2(M, N) <= 2B+1, B = [N^(1/2)], for every M from M_2(N) to N. These are finite sets chosen for each N; an infinite set as in problem 33 may use elements beyond delta N, so the paper bounds neither its liminf nor its limsup and decides neither question.

Results. Read status: claims checked; page numbers are the journal's (pp. 292--300).

• Proposition (p. 292; proof p. 294): with B = [N^{1/k}], B^k-(B-1)^k-1 <= M_k(N) <= (B+1)^k-B^k-1, where M_k(N) is the least M for which some subset of [0,M] completes the kth powers up to N.
• Theorem 1 (p. 293; proof pp. 294--298): there are eps_0(k) > 0 and N_0(k) > 1 such that for N >= N_0 and 3k^2 N^{-1/2k} <= eps <= eps_0, f_k(delta N, N) >= (k-eps)N^{1-1/k} with delta = eps^2/(9k^3).
• Theorem 2 (p. 293; proof p. 298): for all k >= 2 and M_k(N) <= M <= N, f_k(M,N) <= (B+1)^k - B^k with B = [N^{1/k}].
• Corollary (p. 293; proof pp. 298--299): there are 0 < delta_k < 1 and N_k > 1 such that for N >= N_k and M_k/N < delta < delta_k, (k - 3k^{3/2}delta^{1/2})N^{1-1/k} <= f_k(delta N,N) <= k(1+kN^{-1/k})N^{1-1/k}; hence f_k(delta N,N) = (k+o(1))N^{1-1/k} when delta = delta(N) tends to 0 as N tends to infinity.
• Theorem 3 (p. 293; proof p. 299): for fixed delta in (0,1) there are k_0(delta) > 2 and N_k(delta) > 2 such that for k >= k_0(delta) and N >= N_k(delta), f_k(delta N,N) >= C(delta)(k^Delta/log k)(1 - log(1+1/delta)/log k)N^{1-1/k}, where Delta = log(1/delta)/log(1+1/delta) and C(delta) = delta Delta^Delta log(1+1/delta)/(1+Delta)^{1+Delta}.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.