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Statement

For a finite sequence $\{a_i\}$ the $c$-sums are the sums $\sum_{u\le i\le v}a_i$ over all index pairs $1\le u\le v$ (p. 193). The question and the definition, quoted from p. 197: "Prof. Erdős asked the following question in connection with this (personal communication). Is it true that if $\{a_i\}$ is an increasing sequence and

$$a_{i+1}-a_i\le K,\quad K\in\mathbf N,\tag{3.1}$$

then there exist at least two $c$-sums which are equal if $a_i$ is large enough? The answer is yes and we establish this statement in a quantitative form. Let $f(a,K)$ be the largest integer with the following property: There exists an increasing sequence $a=a_1<a_2<\ldots<a_s=f(a,K)$ such that $a_{i+1}-a_i\le K$, $i=1,2,\ldots,s-1$ and all $c$-sums are different."

Theorem 3. "We have $f(a,K)<(a+K/2)e^{K+1}+Ke^{2K+2}$."

As printed on p. 197. The same page prints "It is easy to see that $f(1,1)=2$, $f(2,1)=4$, $f(1,2)=7$, $f(2,2)=10$ and we have seen in the preceding section that $a+(1+o(1))2\sqrt a<f(a,1)<a+(1+o(1))5\sqrt a$ for $a>a_0$" (from Theorem 2, p. 195, on translates of $\{1,\ldots,k\}$), and introduces the theorem as an upper bound showing that $f(a,K)$ exists for all $a$ and $K$ (the sentence prints $f(a,k)$ with a lower-case $k$).

In the problem's notation. Problem 1213 asks for $f(a,K)$ such that an integer sequence $a=a_1<\cdots<a_s$ with $a_s>f(a,K)$ and $a_{i+1}-a_i\le K$ has two distinct intervals $I$, $J$ of indices with $\sum_{i\in I}a_i= \sum_{j\in J}a_j$. Two distinct intervals with equal sums are exactly two equal $c$-sums, overlapping intervals included, so the paper's largest last term of a sequence with all $c$-sums different is the problem's threshold: every such sequence with $a_s>(a+K/2)e^{K+1}+Ke^{2K+2}$ has two equal $c$-sums. The sequence starts exactly at $a$, as in the problem. Since $(a+K/2)e^{K+1}+Ke^{2K+2}\le a\bigl(e^{K+1}+2Ke^{2K+2}\bigr)$ for $a\ge1$, the bound is the site's $f(a,K)\ll ae^{O(K)}$ with explicit constants.

Three filing observations, not review verdicts: the theorem prints the strict inequality while the proof's last line (p. 198) concludes "$f(a,K)\le L$" with $L=e^{K+1}(a+K/2)+Ke^{2K+2}$, and the paper prints no remark on whether the exponential dependence on $K$ is best possible; the site's commentary attributes such a belief to the author, and it has no printed counterpart in this paper; and the printed step from (3.6) to (3.7) fails for large $a$ at every $K$, since with $A=[e^{K+1}]$ the coefficient $A/(\log A-K)$ of $a$ that (3.7) needs exceeds $e^{K+1}$ (for $K=1$, $a=10^4$ and $D=L+1$, $S'-D\approx-74$), while the floor sum $S$ of (3.5), which keeps the slack that (3.6) discards, still exceeds $D$ there ($S-D=47770$ at $D=\lfloor L\rfloor+1$); the conclusion survives, as the Problem 1213 page records.

Source. N. Hegyvári, On consecutive sums in sequences, Acta Math. Hung. 48 (1--2) (1986), 193--200; the question, the definition of $f(a,K)$ and Theorem 3 on printed p. 197 (PDF p. 5 of the publisher scan), its proof on pp. 197--198 (PDF pp. 5--6), read on the page images. The edition is identified in the source digest.

Read depth. Claims checked: the question, the definition, the small values, the $K=1$ estimate and the statement were read clause by clause on the page image on 2026-09-22. The proof (about a page) was read in full on the page images and followed for structure; apart from the passage from (3.6) to (3.7) in the third filing observation, no step, in particular the estimate (3.6) of the block count, was checked. Nothing here is independently reviewed.

Proof pointer

Pages 197--198. Fix $D$ and count blocks $a_{i+1}+\cdots+a_{i+j}$ whose $c$-sum is below $D$. From (3.1), $a_{i+1}\le a+iK$ (display (3.2)), so $a_{i+1}+\cdots+a_{i+j}\le j(a+K/2)+\frac K2(2ij+j^2)$ (display (3.3)), and the block has $c$-sum below $D$ whenever $i\le\frac D{Kj}-\frac{a+K/2}K-\frac j2$ (display (3.4) rearranged). Summing over lengths $j=1,\ldots,A$ gives at least $S=\sum_{j=1}^A\bigl[\frac D{Kj}-\frac{a+K/2}K-\frac j2\bigr]$ such blocks (display (3.5)), and $S>S'=\frac DK\log A-\frac{A(a+K/2)}K-\frac{(A+2)^2}4$ (display (3.6)). If $S'\ge D$ (display (3.7)) two of these blocks have equal $c$-sums, since all of them lie below $D$; (3.7) reads $D(\log A-K)>A(a+K/2)+\frac{K(A+2)^2}4$, and with $A=[e^{K+1}]$ the paper takes it to hold for $D>L=e^{K+1}(a+K/2)+Ke^{2K+2}$, a step that fails for large $a$ (the third filing observation above). Equal $c$-sums then occur among the blocks whose $c$-sum lies in $[1,L]$, and the paper concludes $f(a,K)\le L$.

Dependencies

None outside the paper; the argument is a counting of blocks against the range of their sums. Within the paper, the $K=1$ estimate quoted on p. 197 rests on Theorem 2 (p. 195), whose proof was read for structure only.

Bears on

• Problem 1213: the theorem the site's commentary cites for the affirmative answer, with the exact hypotheses ($a_1=a$, increasing, gaps at most $K$, all $c$-sums over all index pairs distinct) and the explicit bound behind the site's $f(a,K)\ll ae^{O(K)}$; the small values and the $K=1$ estimate of the same page are the paper's only lower bounds.