GPT 6 Astra Ultra attempt on Math-6240
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Contents
Erdős Problem 1186 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
Fix $k\geq2$. Among all red--blue colourings of $[n]$, minimise the number of monochromatic arithmetic progressions of length $k$, counting each progression with positive common difference once. Determine or sharply estimate the optimal asymptotic coefficient $\delta_k$ in the universal lower bound \[\#\{\text{monochromatic }k\text{-term progressions}\}\geq (\delta_k+o(1))n^2.\] An asymptotic description of $\delta_k$ as $k$ grows is requested.CONTEXT AND SCOPE
This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.The page leaves the counting convention implicit; its displayed random-colouring bound 1/$(k-1)2^k$ identifies the convention used here.
ATTACK PLAN AND WORK
For fixed $k\ge2$, the number of $k$-term progressions in $[n]$, with positive common difference counted once, is \[T_k(n)=\sum_{d=1}^{\lfloor(n-1)/(k-1)\rfloor}(n-(k-1)d) =\frac{n^2}{2(k-1)}+O_k(n).\tag{1}\] In an independent fair coloring each progression is monochromatic with probability $2^{1-k}$. Linearity of expectation, which requires no independence between different progressions, therefore gives a coloring with at most \[\frac{n^2}{(k-1)2^k}+O_k(n)\] monochromatic progressions. In particular the optimal coefficient satisfies $\delta_k\le1/((k-1)2^k)$.For $k=2$ this calculation is sharp. If the color classes have sizes $a,n-a$, every unordered pair is a 2-term progression, and their monochromatic count is \[\binom a2+\binom{n-a}2 =\left(a-\frac n2\right)^2+\frac{n^2}{4}-\frac n2.\] Balanced classes attain the minimum up to integer rounding, so $\delta_2=1/4$. The square term proves the lower bound directly; averaging alone would not.