# Upper bound for the number of primes in a series of n consecutive odd positive integers (all greater than 3)

Apart from 3, 5, 7, there cannot be three consecutive odd numbers which are all primes, as is well known. I wonder how this fact* can be used to calculate the upper bound in the title for any n.

*: Whence the condition that the integers be greater than 3

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In any $n$ consecutive odd integers, there are $\lfloor n/3\rfloor$ disjoint triples of $3$ consecutive odd integers, so at least that many of the integers are divisible by $3$. Therefore (if all the integers are greater than $3$) there are at least $\lfloor n/3 \rfloor$ composites and at most $n - \lfloor n/3 \rfloor = \lceil 2n/3 \rceil$ primes.

EDIT: For computations involving $\lfloor \ldots \rfloor$ and $\lceil \ldots\rceil$, it's often best to "unwind" the definition. $x = \lfloor n/3 \rfloor$ means that $x$ is an integer with $x \le n/3 < x + 1$. Then $n - \lfloor n/3 \rfloor = n - x$ is an integer with $n - x - 1 < 2n/3 \le n - x$. And that's the definition of $\lceil 2n/3 \rceil$.

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Many thanks for your reply. However, what if n does not divide 3? Consider, for example, n=4. – user82365 Jun 13 '13 at 22:51
$\lfloor 4/3 \rfloor = 1$. Look up "floor" and "ceiling". – Robert Israel Jun 14 '13 at 6:07
Done! Now I see, thanks. Would you please just add one reference as to how to compute the subtraction at the end? Specifically, as to why the result is a ceiling. – user82365 Jun 14 '13 at 7:19
Perfect! Thanks again – user82365 Jun 14 '13 at 8:53

You can reduce the upper bound in many cases by proving that of the remaining numbers, some have to be divisible by $5, 7, 11$ etc. The first example is $n = 8$. You can have $6$ numbers not divisible by $3$, but only if the numbers are $6k+5$ to $6k+19$, where $6k+5$, $6k+7$, $6k+11$, $6k+13$, $6k+17, 6k+19$ are not divisible by $3$. But no matter how you choose k, one of these is divisible by $5$, so there can be only $5$ instead of $6$ primes. For large $n$, finding the upper bound is quite difficult.

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