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MAXIMUM GCD AMONG PAIRS OF RANDOM INTEGERS

机译:随机整数成对的最大GCD

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摘要

Fix α > 0, and sample N integers uniformly at random from ! 1, 2, . . . , " eαN #$ . Given η > 0, the probability that the maximum of the pairwise GCDs lies between N2?η and N2+η converges to 1 as N → ∞. More precise estimates are obtained. This is a Birthday Problem: two of the random integers are likely to share some prime factor of order N2% log(N). The proof generalizes to any arithmetical semigroup where a suitable form of the Prime Number Theorem is valid.
机译:修复α> 0,并随机地均匀地样本。 1,2,。 。 。 ,“eαn#$。给定η> 0,成对gcds的最大值在n2Δη和n2 +η之间会聚到1的概率为n→∞。获得更精确的估计。这是生日问题:二随机整数可能分享某些命令N2%Log(n)的主要因子。证明概括了任何算术半群,其中素数定理的合适形式是有效的。

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