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Sylvester-Schur says: "if $n \ge 2k$, then there is a number in the list $n − k + 1, n − k + 2,$ ... $, n$ divisible by a prime $p > k$."

Shouldn't it also be true that if $n \ge k$, then there is a number in the list $n + 1, n + 2, $ ... $, n+k$ divisible by a prime $p > k$.

Does anyone know of a theorem that comes close to establish this or why the stronger claim is not true?

Thanks very much,

-Larry

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Thanks for noticing! I just fixed it. :-). –  Larry Freeman Nov 8 '12 at 16:32

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up vote 2 down vote accepted

Yes, this is true.

Indeed, by a result of Denis Hanson (Canad. Math. Bull., 1973) the product (I use notation to match the papers not the question) $$\Delta(n,k)= n (n+1) \dots (n+k-1)$$ for $n \ge k$ is divisible by a prime of size greater $3k/2$ with only the exception of $3\cdot 4$ , $8 \cdot 9$ and $6\cdot 7 \dots 10$.

There are also further results in this direction. For example Laishram and Shorey (Acta Arith., 2005) proved that the largest prime divisor of $\Delta(n,k)$ is strictly greater $1.97k$ if $n \gt k+13$ and $2k$ if in addition $n> (279/262) k$

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Based on the exceptions you show in the Hanson result, I think the prime has to be strictly greater than 3k/2. Gerhard "Ask Me About Strict Interpretation" Paseman, 2012.11.08 –  Gerhard Paseman Nov 8 '12 at 21:37
    
Yes, you should be right. Thanks! I cannot recheck at the moment, but the exceptions would not be exceptions and in any case it only changes anything for k=2, where later on this is certainly fine, so yes 'greater' not 'at least'. –  quid Nov 8 '12 at 22:35

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