This challenge was the last before an extended summer break. We left off unsure whether or not a sequence reached infinity at a finite point. Well, it does, and almost certainly at 23. But why? The solution is below.
Let us arrange the set of natural numbers into a sequence T. T begins with 1. For any Tn, Tn+1 is the lowest number prime connected to n that is not in the sequence T. If there exists none, then look to Tn-1, and repeat this process. If there are no more prime numbers connected to any previous numbers, then continue with the lowest natural number not yet in the set.
Within these constraints, a is always maximized, and n minimized for the value of a. This sequence is then: 1, 2, 4, 3, 6, …, 23. Now, what happens if this set is infinite? Well, some numbers would be skipped. Specifically, any dead end of connections not immediately followed would never be. This sequence is actually the best way to compute these sets. But computing to a high amount isn’t proof. But this is still helpful.
In order for there to be no infinite sets, there must be an infinite amount of finite sets. For each prime or the form 6n±1, or all those greater than 3, there are at least 3 numbers in its set. For 6n+1, they are 6n+1, 12n+2, and 12n. For 6n-1, they are 6n-1, 12n-2, and 12n-4. When n is even, both extend to 12n-3 and 12n-6. This proves that twin primes like 11 and 13, but not necessarily 17 and 19, must connect. This matches experimentation. Consider, in this sequence, that 2, 3, 5, 7, 9, 11, 17, 19, and 23 all are the first of their sets. Is 9 special? Yes, for any prime multiple of 3, only 1 number is connected to it, which is below. But 9 is alone, and so can start, and end, its set. But since there is no infinite occurrence of this, then there must be an infinite number of disconnected primes. This is impossible.
The reason for this, at last, is simple: an infinite number of sets that have a prime lowest number may not leave any composite spaces. That’s where the sequence comes in: since the first element the lowest and prime, a prime’s set must cover each number up to the next prime. But it must also contain those 2 additional numbers mentioned earlier. The next set must be unable to reach those, but able to reach all others. Here probability does the trick: as one goes through primes, the probability of this dwindles on the order below. Since this does not approach infinity, the weak law of large numbers tells us there must be an infinite set, and since 23’s grows so much so quickly, it is almost certainly the answer. Though there is a tiny probability of otherwise, and there could be a great number of very large sets afterwards, there must be a large one eventually, if not immediately.
\operatorname \mathcal 0 {\left({n\over{2\choose{\frac n{\ln n}}}\ln n}\right)}Thus concludes Preposterous Primes, and zacharycormack.net‘s extended summer break.