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Larry Freeman
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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?

Once again, apologies if this is a naive question.

Thanks very much,

-Larry

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?

Once again, apologies if this is a naive question.

Thanks very much,

-Larry

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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Larry Freeman
  • 1k
  • 1
  • 7
  • 19

Are there any theorems about a prime $p > k$ in a sequence stronger than Sylvester-Schur?

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?

Once again, apologies if this is a naive question.

Thanks very much,

-Larry