Erdős Problem #1211
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Erdős problems openaigpt 6 astra ultra 65% cataloguedSolvedSolvedStatement
For a partition $\mathbb N=A\sqcup B$, let $S(A)$ and $S(B)$ be the sets of nonempty sums of distinct elements of the corresponding part. How small can \[ \max\left(\overline\delta_{\log}(S(A)), \overline\delta_{\log}(S(B))\right) \] be? Conlon--Fox--Pham proved that the optimum is $(2+\sqrt3)/4$. We reconstruct the sharp upper-bound construction, including its logarithmic-density calculation, and prove the elementary universal lower bound $1/2$. The stronger universal lower bound matching the construction remains unresolved in this attempt.
Restatement from Math-6265-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
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GPT 6 Astra Ultra attempt on Math-6265
Erdős Problem 1211: the extremal logarithmic block construction
Statement and scope
For a partition $\mathbb N=A\sqcup B$, let $S(A)$ and $S(B)$ be the sets of nonempty sums of distinct elements of the corresponding part. How small can \[ \max\left(\overline\delta_{\log}(S(A)), \overline\delta_{\log}(S(B))\right) \] be? Conlon--Fox--Pham proved that the optimum is $(2+\sqrt3)/4$. We reconstruct the sharp upper-bound construction, including its logarithmic-density calculation, and prove the elementary universal lower bound $1/2$. The stronger universal lower bound matching the construction remains unresolved in this attempt.A universal elementary lower bound
Singletons are allowed as sums, so $A\subseteq S(A)$ and $B\subseteq S(B)$. Since the two parts partition the positive integers, \[ \sum_{\substack{n\in A\\n\le x}}\frac1n+ \sum_{\substack{n\in B\\n\le x}}\frac1n =\log x+O(1). \] Taking upper limits proves that at least one of $S(A),S(B)$ has upper logarithmic density at least $1/2$. This gives no justification for the much larger sharp constant.Alternating blocks on the logarithm-of-logarithm scale
Fix $b>2$. Assign the integers in \[ [e^{b^{2j}},e^{b^{2j+1}})\quad(j=0,1,\ldots) \] to $A$, and those in the intervening blocks to $B$. Assign the finitely many initial integers arbitrarily. Let $R_j=e^{b^{2j+1}}$ be the upper endpoint of the $j$th $A$-block and $L_{j+1}=e^{b^{2j+2}}$ the lower endpoint of the next one.Any element of $S(A)$ below $L_{j+1}$ uses only summands below $R_j$. Its value is at most the sum of all positive integers below $R_j$, which is less than $R_j^2$ for large $j$. Hence $S(A)$ avoids the entire integer interval \[ (R_j^2,L_{j+1}) =\left(e^{2b^{2j+1}},e^{b^{2j+2}}\right). \tag{1} \] The same argument gives the shifted family of gaps for $S(B)$. All earlier blocks are automatically accounted for by bounding their total by the sum of all integers below $R_j$.
Computing an upper density uniformly over the cutoff
For a set of integer gaps $(e^u,e^v)$, the harmonic weight removed is $v-u+O(e^{-u})$. The errors over the gap family $1$ have bounded sum. Thus, on the variable $t=\log x$, it suffices to examine the length proportion removed by intervals \[ (u_j,v_j)=(2b^{2j+1},b^{2j+2}). \] Their lengths are $(b-2)b^{2j+1}$, and $u_{j+1}/u_j=b^2$.The removed-length proportion decreases between gaps and increases inside a gap: during a gap its numerator is $t$ minus a fixed nonnegative covered length. Its asymptotic minimum therefore occurs immediately before a gap, at $t=u_j$. At that point the earlier removed length is \[ \sum_{h<j}(v_h-u_h) =(b-2)b\,\frac{b^{2j}-1}{b^2-1}, \] up to a fixed initial adjustment. Division by $u_j=2b^{2j+1}$ gives the limit \[ \frac{b-2}{2(b^2-1)}. \] The uniform-over-cutoffs conclusion is \[ \overline\delta_{\log}(S(A)),\ \overline\delta_{\log}(S(B)) \le D(b):=1-\frac{b-2}{2(b^2-1)}. \tag{2} \] For $B$, multiplying all gap endpoints by $b$ leaves this limiting proportion unchanged. Integer rounding and finite initial choices contribute only $O(1)/\log x$.