Prove that there always exists a sequence of consecutive composite integers of length $n$, for all $n$.
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You mean arithmetic sequence? – Pedro Oct 09 '13 at 16:27
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If this is just about arithmetic sequences: All terms of the sequence $100n+10$ are composite (I deliberately avoided $0$) – Hagen von Eitzen Oct 09 '13 at 16:35
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I meant to say consecutive numbers – Saumya Oct 09 '13 at 16:43
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/APOSTOL/Ch4/Ex5 – nature1729 Aug 29 '14 at 10:45
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Consider the sequence $(n+1)!+2,(n+1)!+3,\cdots ,(n+1)!+n+1$. Each of these numbers is a composiite number since $(n+1)!+2$ is divisible by 2,$(n+1)!+3$ is divisible by 3 and so on.
Shaswata
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