GPT 6 Astra Ultra attempt on Math-6254
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Contents
Erdős Problem 1200 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
Does an absolute constant $C>0$ exist such that, for all sufficiently large $x$, one can select distinct primes $p_1<\cdots<p_k<x$ and residues $a_i\pmod{p_i}$ satisfying \[\sum_{i=1}^k\frac1{p_i}<C,\qquad [1,x)\cap\mathbb Z\subseteq\bigcup_{i=1}^k(a_i+p_i\mathbb Z)?\]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 says every integer n<x; the finite positive interval is the intended interpretation of this covering question.
ATTACK PLAN AND WORK
Let $M=\lceil x\rceil-1$, so the interval to be covered is $\{1,\ldots,M\}$. One class modulo $p_i$ covers at most $M/p_i+1$ of these integers. Therefore every proposed cover satisfies \[1\le\sum_{i=1}^k\frac1{p_i}+\frac kM.\tag{1}\] In particular a construction using $o(M)$ primes needs reciprocal mass at least $1-o(1)$. The usual fact $\pi(x)=o(x)$ gives this condition for the available primes as well; it is only a necessary lower threshold, not an impossibility result.There is also an exact greedy estimate for any prescribed distinct primes. If $U$ integers remain uncovered, averaging over the $p$ residue classes finds one covering at least $U/p$ of them. Choosing such a class successively gives \[U_{\rm final}\le M\prod_i(1-1/p_i).\tag{2}\] No independence of the uncovered residue patterns is assumed in this averaging. However $2$ need not be below one when the reciprocal sum is bounded: for large selected primes the product is of constant scale, leaving an upper bound of order $M$. Failure of this bound to prove coverage is not a lower bound on what a better choice of residues can cover.