Being given $a \in \mathbb{N}$, find the smallest $N$ such that $\{ N, N+1, N+2, N+3, N+4, ...., N+a\}$ are all(consecutive) composite numbers.
For example for $a=2$, $N=8$ ; for $a=4$, $N=24$.
I am looking for a general formula.
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thats not a question – supinf Oct 20 '16 at 07:59
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No general formula known. But an example of such $N$ is known. – Sarvesh Ravichandran Iyer Oct 20 '16 at 08:00
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Take $N=m!+2$ for some and that $m> a$ – arberavdullahu Oct 20 '16 at 08:26
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1The numbers are tabulated at http://oeis.org/A008950 where there are some references (but no formulas, and as others have said no one knows any formulas). – Gerry Myerson Oct 20 '16 at 09:12
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Great! Thank you Gerry Myerson. You helped me a lot. – Amir Oct 20 '16 at 09:16
1 Answers
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Let us consider the natural, classical, different, question: given any $a$, is it always possible to find a sequence of $a+1$ consecutive composite numbers ?
The answer is: it suffices to take $M=(a+1)!$ (factorial of $(a+1)$). Why that?
Thus, this constitutes a partial answer to your question: your $N$ is at most $(a+1)!+2$.
Jean Marie
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Yes. It is obvious that N=< (a+1)!+2. I am looking for the minimum. Formula or pattern? – Amir Oct 20 '16 at 08:57
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I couldn't guess you know this result because you don't speak about it. Besides, as астон вілла олоф мэллбэрг, I am fairly certain that no formula exist for this minimum $N$. – Jean Marie Oct 20 '16 at 09:07
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I meant no offence. I appreciate your answer and comment. :) And thanks to астон вілла олоф мэллбэрг. – Amir Oct 20 '16 at 09:12