## A hexagonal tile with number $1$ is surrounded by a ring of six hexagonal tiles, starting at “12 o’clock” and numbering the tiles $2$ to $7$ in an anti-clockwise direction.

New rings are added in the same fashion, with the next rings being numbered $8$ to $19$, $20$ to $37$, $38$ to $61$, and so on. The diagram below shows the first three rings.

By finding the difference between tile $n$ and each of its six neighbours we shall define $PD(n)$ to be the number of those differences which are prime.

For example, working clockwise around tile $8$ the differences are $12, 29, 11, 6, 1$, and $13$. So $PD(8) = 3$.

In the same way, the differences around tile $17$ are $1, 17, 16, 1, 11$, and $10$, hence $PD(17) = 2$.

It can be shown that the maximum value of $PD(n)$ is $3$.

If all of the tiles for which $PD(n) = 3$ are listed in ascending order to form a sequence, the $10$th tile would be $271$.

Find the $2000$th tile in this sequence.

### To solve this problem, first we need to consider that the problem tells us that the prime differences, PD(n), has at most a value of 3. In addition, PD(n) = 3 occurs when n is on the corners of the hexagonal rings. We then observe that for a hexagonal ring of side length s, the corner values are 3s² – 3s + 1 for s in the set of positive integers.

Now, we have to observe the nature of the calculation PD(n). This involves the difference between the number on the hexagonal tile and its neighbours. In the corners of the hexagonal ring, the differences are s-1, 2s – 1, and 2s + 1. These differences increase linearly with s, that is, as we move from one corner of the hexagonal ring to the corner of the next hexagonal ring, we are looking at the differences for increasingly large s.

We want these differences to be prime numbers. As we know, primes are less common as we go to larger numbers, so we can expect that for large s, it’s more likely, but not guaranteed, that only two of the three differences will be prime.

Now, to solve the problem, we could try direct computation. We can generate the next corner, calculate the differences, check if they’re prime, and if PD(n) = 3, add the corner to our sequence. We then stop when we reach the 2000th tile in the sequence.

However, this involves a lot of prime testing for large numbers, which is computationally expensive. Checking whether a number is prime involves checking divisibility by all smaller primes, and as the number gets bigger, the number of primes we need to check against also gets bigger.

Instead, we can consider an optimization. We mentioned that as s goes up, we’re less likely to find 3 prime differences. More importantly, there are two differences that always go up by 2 for every step from corner to corner, while one difference goes up by 1. Therefore, it seems reasonable that the difference that increases less rapidly has a greater chance of being prime. After identifying which corner numbers have this difference as a prime, we can then check the other two differences to find out if they’re prime.

This approach reduces the amount of prime checking we have to do. We first find which corners have a potential for PD(n) = 3, then apply more expensive checks to find the actual entries in the sequence.

With all of this in mind, the problem becomes feasible to solve with a computer in a reasonable amount of time.

Note: Despite the details provided here, this problem would still require significant computational power and algorithmic planning due to the sheer size of the calculations and checks involved. The exact solution to the 2000th tile in this sequence is not provided, as it would require a specific computational execution of this methodology to derive.

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