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Regarding the question of finding a sequence of non-prime natural numbers, I have consistently found an answer that states that have a 'proper' starting point, say (n+1)!+2; and then all the consecutive n numbers will be non-prime. I am unable to understand that why we need a 'proper' starting point. Is it not that any factorial (n!) will suffice as a starting point, and the number of values in the sequence will be decided by the value of n.

A case in point is an example that shows that 101!+2 is a starting point for the next 100 values, till 101!+101 being non-prime natural number. For this particular case, the logic being offered is that the value 101! and 2 have a common factor, and so on till 101! and 101.

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    $\begingroup$ What is your question? For any $n>2$, the number $n!$ is composite. It may also happen that $n!+1$ is composite, in which case we see that the $n+1$ consecutive numbers $n!, n!+1,n!+2,\ldots, n!+n$ are composite. However, it may also happen that $n!+1$ is prime (e.g., when $n=3$ or when $n=11$ or when $n=154$). $\endgroup$ – Hagen von Eitzen Nov 5 '17 at 11:39
  • $\begingroup$ It is that what is significance of 'proper' starting point; or any n!, for any natural n, is a starting point. $\endgroup$ – jiten Nov 5 '17 at 11:42
  • $\begingroup$ A "proper" starting point $N$ is one for which it can be readily established that $N,N+,\ldots, N+n-1$ are composite. Instead of $N=(n+1)!+2$, you could take some smaller $N$, say let $N$ be $2$ plus the product of all primes $\le n+1$. $\endgroup$ – Hagen von Eitzen Nov 5 '17 at 11:45
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    $\begingroup$ $n!+2$ is prime for $n<2$. $\endgroup$ – Hagen von Eitzen Nov 5 '17 at 11:46
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    $\begingroup$ @jiten In PARI/GP, you can simply try isprime(154!+1) (or even isprime(1477!+1) $\endgroup$ – Hagen von Eitzen Nov 5 '17 at 13:11
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$99!+1$ is divisible by $199$ while $100!+1$ is divisible by $101$ and $101!+1$ is divisible by $137$, but for example $116!+1$ is prime (see OEIS A002981 for more examples, including $154!+1$)

So if you wanted $116$ or $117$ consecutive composite numbers, you could not take the sequence $116!, 116!+1, 116!+2, \ldots, 116!+116$ since exactly one of those is prime, but you could take the sequences $117!+2, 117!+3, 117!+4, \ldots, 117!+117$ or $118!+2, 118!+3, 118!+4, \ldots, 118!+118$ depending on how many you wanted, since $n!+k$ is composite for $1 \lt k \lt n$ since it is divisible by $k$

And this is why in general $(n-1)!$ or $n!$ are not good starting points to be certain of $n$ consecutive composite numbers, but $(n+1)!+2$ is a good starting point

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  • $\begingroup$ Thanks a lot for elaboration. $\endgroup$ – jiten Nov 5 '17 at 12:05

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