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The prime factors of consecutive integers. (English) Zbl 0128.04002

If \(f(k)\) denotes the least integer such that each product of \(f(k)\) consecutive integers all greater than \(k\) has a prime factor greater than \(k\), then it is clear that if \(k\) is composite, the value of \(f(k)\) is that of \(f(p)\) where \(p\) is the largest prime smaller than \(k\). It is known [cf. P. Erdős, Nieuw Arch. Wiskd., III. Ser. 3, 124–128 (1955; Zbl 0065.27605)] that \(f(2) = 2\), \(f(3) = 3\), \(f(5) = f(7) = 4\) and that \(f(13) > 6\).
In this paper a machine method is determined to give any \(f(k)\). In particular, this method yields \(f(11) = 4\), \(f(13) = f(17) = f(19) = f(23) = f(29) = f(31) = f(37) = 6\) and \(f(41) = 7\).
Reviewer: W. R. Utz

MSC:

11A51 Factorization; primality
11Y05 Factorization

Citations:

Zbl 0065.27605
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