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For any full list of the primes up to the $n$th prime: $P = \{2, 3,5,\dots, p_n\}$, any natural number $q$ such that $ p_n \lt q \lt p_{n+1}^2$ that is not sieved by a prime in $P$ is also a prime. So to count such primes $q$, it's $f(2) = \Delta - [\frac{\Delta}{2}]$ between $2$ and $3\cdot 3$ there are $7 - [7/2] = 3$ primes namely $3,5,7$, where $\Delta = p_{n+1}^2 - p_n$, and $[\cdot]$ is the ceiling function. Similarly with $\Delta = 5^2 - 3 = 22$ we have $f(3) = \Delta - [\Delta/2] - [\Delta/3] + [\Delta/(2\cdot 3)] = 22 - 11 - 8 + 4 = 7 = \sharp \{ 5,7,11,13,17,19, 23\} = \sharp\{q: 3 \lt q \lt 5^2\}$. So given $7$ primes and $5^2 - 3 - 1 = 21$ slots (counting all possible numbers between $p_n = 3$ and $p_{n+1}^2 = 5\cdot 5$ and not including the bounds). If there were no twin primes in that span then we would be able to "place a prime" every $4$ slots or greater which constrains us to no more than $\lfloor 21 / 4\rfloor = 5$ primes. Thus there must exist at least two pairs of twin primes ($7 - 5 = 2$) in that span.

Could this counting method be applied to general $n$ and so help prove the twin prime conjecture?

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    $\begingroup$ I don't think so. The key piece appears to be that if there are no twin primes then primes must be separated by 4 or more, which leaves too little room for all the primes that have to be in the interval. But the problem is, the density of primes $\le n$ decreases as $\frac{1}{\log n}$, so for sufficiently high $n$, there is no problem with them being separated by at least $4$. Still, it might be useful in some other way. $\endgroup$ – Paul Sinclair Aug 30 '15 at 19:13

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