GPT 6 Astra Ultra attempt on Math-779
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Erdős Problem #357: distinct consecutive sums
1. Statement and scope
Let $f(n)$ be the largest $k$ for which some $1\le a_1<\cdots<a_k\le n$ has all sums $\sum_{i=u}^v a_i$, $1\le u\le v\le k$, distinct. How does $f(n)$ grow, and is $f(n)=o(n)$? Related questions ask whether an infinite such sequence has density zero and whether its reciprocal sum converges. We prove explicit lower and upper bounds and an infinite-sequence growth obstruction. The sublinear upper bound and reciprocal-sum questions remain unresolved.2. An explicit lower bound of order $2\sqrt n$
Put $m=\lfloor\sqrt n\rfloor$, $k=2m-1$, and take the last $k$ integers up to $n$, namely $a_i=n-k+i$. These are positive since $m^2\ge2m-1$. Sums of a fixed length $r$ strictly increase with the starting index. Their minimum and maximum are \[ L_r=r(n-k)+r(r+1)/2,\qquad U_r=rn-r(r-1)/2. \] For $1\le r<k$, \[ L_{r+1}-U_r =n-(r+1)(k-1)+r^2 =(r-m+1)^2+n-m^2+1\ge1. \] Thus the ranges of different lengths are disjoint, proving \[ f(n)\ge2\lfloor\sqrt n\rfloor-1. \] This is the familiar interval construction also checked in the earlier GPT Pro 5.2 attempt.3. An elementary upper bound with controlled error
Write $S=\sum_{i=1}^k a_i$. For $1\le h\le k$, there are \[ M=\sum_{r=1}^h(k-r+1)=hk-h(h-1)/2 \] short consecutive sums of lengths at most $h$. They are distinct positive integers, so their total is at least $M(M+1)/2$. Each $a_i$ occurs in at most $r$ sums of length $r$, so the total is at most $h(h+1)S/2$. Therefore \[ M^2\le h(h+1)S,\qquad (k-h/2)^2\le(1+1/h)S. \tag{1} \] Choose $h=\lfloor\sqrt k\rfloor$. If $S\ge k^2$ there is nothing to prove; otherwise $1$ implies \[ k^2\le S+kh+k^2/h=S+O(k^{3/2}). \] The individual $a_i$ are distinct and bounded by $n$, giving $S\le kn-k(k-1)/2$. Combining these inequalities yields \[ f(n)\le\frac23 n+O(\sqrt n). \tag{2} \] The argument even permits a nonmonotone ordering, because distinct length-one sums still force distinct entries. It reconstructs a classical weaker upper bound; it is not asserted to be the best published constant.4. A quantitative obstruction for an infinite sequence
For any infinite increasing sequence with the stated property, \[ \limsup_{j\to\infty}\frac{a_j}{j\log j}\ge\frac12. \tag{3} \] Suppose instead that $a_j\le c j\log j$ for all sufficiently large $j$, with $c<1/2$. Choose a fixed small $\lambda>0$ so that $1/(c(1+\lambda))>2$. For large starting indices $j\le N/(1+\lambda)$ and $1\le r\le\lfloor\lambda j\rfloor$, monotonicity gives \[ \sum_{i=j}^{j+r-1}a_i \le c(1+\lambda)rj\log((1+\lambda)j). \] All these sums are distinct positive integers at most $S_N=a_1+\cdots+a_N=O(N^2\log N)$. Their reciprocal sum is therefore at most the harmonic sum $H_{S_N}=(2+o(1))\log N$. On the other hand it is at least \[ \frac1{c(1+\lambda)} \sum_{j_0\le j\le N/(1+\lambda)} \frac{H_{\lfloor\lambda j\rfloor}}{j\log((1+\lambda)j)} =\left(\frac1{c(1+\lambda)}+o(1)\right)\log N. \] The last equality follows from $H_{\lfloor\lambda j\rfloor}=\log j+O_\lambda(1)$; the accumulated error is $O_\lambda(\log\log N)$. The two bounds contradict the choice of $\lambda$, proving $3$. In particular $j/a_j$ tends to zero along a subsequence, so the set of terms has lower asymptotic density zero.5. Assessment and references
The interval construction and $2$ leave a large gap between $\sqrt n$ and $n$. Statement $3$ concerns a subsequence in one fixed infinite example and does not yield a uniform bound for all finite examples. Lower density zero is not density zero; nor does $3$ imply convergence of $\sum1/a_j$.Source: https://www.erdosproblems.com/357, checked 2026-09-25, which credits Erdős--Harzheim for the question and records the averaging obstruction and Hegyvári's bounds. The prior GPT Pro 5.2 constructions were checked and retained with explicit attribution. The arguments here use no unproved external theorem, and no novelty or formal Lean verification is claimed.