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Statement

Setting (p. 1, Section 1, repeating the abstract): $k\ge2$ is an integer, $N$ is fixed, and $A^N$ is a set of non-negative integers such that "for all integer $n\le N$, $n$ can be written as $n=a+b^k$, $a\in A^N$, $b$ a positive integer." Since $a\ge0$ and $b\ge1$, only $n\ge1$ can be so written; the paper's Lemma 1 sums over $1\le n\le N$, and the condition is read for those $n$.

Theorem 1 (p. 1, quoted).

$$\lvert A^N\rvert\ge N^{1-\frac1k}\Bigl\{\frac{1}{\Gamma(2-\frac1k)\,\Gamma(1+\frac1k)}+o(1)\Bigr\}.$$

The $o(1)$ is as $N\to\infty$; the proof (pp. 2--5) establishes it in the form

$$\liminf_{N\to\infty}\frac{\lvert A^N\rvert}{N^{1-\frac1k}}\ge\frac{1}{\Gamma(2-\frac1k)\,\Gamma(1+\frac1k)},$$

for any choice of the sets $A^N$ (p. 5). The paper states that the theorem improves a result of Balasubramanian (J. Number Theory 29 (1988), 10--12), and its table on p. 6 compares the constant with Donagi and Herzog's $1+\frac{k-1}{2k^2}$ and Balasubramanian's $(2-\frac2{k+1})^{1/k}$.

The case $k=2$ (evaluated here; the paper prints no numerical value). Since $\Gamma(\tfrac32)=\sqrt\pi/2$, the constant at $k=2$ is $1/\Gamma(\tfrac32)^2=4/\pi=1.2732\ldots$.

Remarks in Section 3 (pp. 6--8). After a table of constants (Observation 1, p. 6), the paper adds two observations, neither labelled as a theorem.

• Sharpness under a hypothesis (Observation 2, pp. 6--7). With $r(n)$ the number of representations $n=a+b^k$, $a\in A$, the paper says that if for each $N$ some $A^N$ has $\sum_{n=1}^N r(n)=N+o(N)$, then Theorem 1 is best possible, and argues that such sets would have liminf at most the theorem's constant. It expects that hypothesis to be false, conjectures $\sum_{n\le N}r(n)\ge kN+o(N)$, and poses as an open problem to find, for each $k$, a constant $c_k>1$ with $\sum_{n\le N}r(n)\ge c_kN+o(N)$. In the displayed integral of that argument (p. 7) the factor $(1-x)$ carries the exponent $\frac1k$; the stated conclusion needs the exponent $\frac1k-1$, the derivative's, since $\int_0^1\frac1k(1-x)^{\frac1k-1}x^{1-\frac1k},dx =\Gamma(2-\frac1k)\Gamma(1+\frac1k)$ (an observation of this page).
• Other sequences (Observation 3, pp. 7--8). The proof uses no arithmetic property of the $k$th powers, and the paper states that it extends to sequences $b_n=\beta n^\gamma+o(n^\gamma)$ with $\beta>0$, $\gamma>1$, giving $\liminf_{N\to\infty}\lvert A\rvert/N^{1-\frac1\gamma}\ge \beta^{\frac1\gamma}/(\Gamma(2-\frac1\gamma)\Gamma(1+\frac1\gamma))$. No proof of the extension is written out.

Source. J. Cilleruelo, The additive completion of $k$th-powers, J. Number Theory 44 (1993), no. 3, 237--243, doi:10.1006/jnth.1993.1049, read in the author-typeset manuscript identified on the source card, whose pages are numbered 1 to 8 and carry no journal pagination: the setting and Theorem 1 on p. 1, the proof in Section 2 on pp. 2--5, the observations in Section 3 on pp. 6--8.

Read depth. Claims checked: the setting, the statement and the Section 3 remarks were read clause by clause on the page images. The proof was read but not checked step by step, and the limit $\lim_{\alpha\to1}c_\alpha$, which the paper leaves to the reader, was not computed here. Nothing here is independently reviewed.

Proof pointer

Pp. 1--5. Two lemmas set up a weighted count. By Lemma 1 (p. 1, attributed to Balasubramanian), for any $f\ge0$ the sum of $f(a+b^k)$ over the representations $a+b^k\le N$ is at least $\sum_{n=1}^Nf(n)$, because every $n\le N$ has at least one representation. Lemma 2 (p. 2) turns both sides into integrals for a weight $f(x)=g(x/N)$, which gives $\sum_{a\in A}h(a/N)\ge N^{1-1/k}\int_0^1g+O(N^{1-2/k})$ (the paper's (2)) with $h$ the profile of Lemma 2. For weights whose profile rises to a single interior maximum at $y_0$ (conditions (i)--(v), p. 2), the paper splits $A^N$ into blocks below $Ny_0$ and the rest, applies partial summation, and feeds a known lower bound $c_0$ for the liminf back in on the smaller ranges; this yields an improved bound $c_1$, and iterating gives a bound for each admissible $g$ in closed form (p. 4). The weights $g_\alpha(x)=\max(x-\alpha,0)$, $\alpha<1$, have explicit profiles (p. 5), and letting $\alpha\to1$ gives the gamma-function constant.

Bears on

• Problem 33: the problem asks, for a set $A$ such that every large integer is $n^2+a$ with $a\in A$ and $n\ge0$, whether $\liminf\lvert A\cap\{1,\ldots,N\}\rvert/N^{1/2}>1$. At $k=2$ the theorem gives the constant $4/\pi>1$, but it asks for $b\ge1$, while the problem also admits $n=0$. The problem's claim page for this paper extends the proof to $b=0$ through the error term of Lemma 2 and so credits the theorem with the answer yes to the liminf question; the paper itself states only $b\ge1$. The theorem is a lower bound and does not determine the smallest possible limsup that the problem's first question asks for.