Lol, wut?

I had the opportunity to catch up with my old friend, Chai, earlier this week. It has been a while since we talked and the topic of aging and birthdays came up.

Chai is turning 32, or \(2^5\) years old(!!), later on this year. This is the (reasonably speaking) penultimate \(2^n\)th of age in most of our lives: while an uncomfortable thought, it is highly unlikely anyone reading this article in the year 2022 will live to the age of \(2^7\)). (BTW: Chai if you’re reading this (and Date.now() < 1670302800000): happy early birthday, fam! 🎉🎉)

During that conversation, something occurred to me: the numbers \([33, 34, 35]\) are consecutive (distinct) semiprimes!

A quick refresher:

semiprime: a natural number that is the product of exactly two prime numbers

And:

distinct semiprime: a semiprime that is not a perfect square (eg: 25 may count as a semiprime but it is not a distinct semiprime)

Observing that \([33, 34, 35]\) were consecutive is pretty neat: it occurred to me that this particular set was probably special (or at the very least, interesting to me).

After some reflection, I am certain that:

1. The longest possible chain of consecutive, distinct semiprimes is three. For fun, I will refer to this as a 🥪 semiprime sandwich 🥪
2. The first occurance of a semiprime sandwich starts at 33 (so 33, 34, 35 is the first semiprime sandwich).

(SIDEBAR| This my friends, makes the 30s a delightfully special decade of our lives - the next time we may hope to encounter a semiprime sandwich age range would be from age 93!)

(FWIW, this curious property is not present on the wikipedia page on semiprimes, which leads me to believe that either this observation is obvious, inconsequential, or both. Still, it was fun to think about and for me at least, a source of delight during an otherwise stressful week. Anyways, onwards with this post!)

Justification

Note, I am not labeling this a proof because my reasoning below will be pretty informal.

Suppose we have some number sequence \(n-1\), \(n\), and \(n+1\) such that they are all distinct semiprimes. (It goes without saying that they are all also consecutive).

We have two options to consider:

• \(n\) could be odd
• \(n\) could be even

A (possible) semiprime 🍞\(n\)🍞 where \(n\) is odd.

If \(n\) is odd and a semiprime, we know that its two prime factors must also be odd. (eg: if \(n == 35\) then its factors are \(5\) and \(7\)).

Also, because \(n\) is odd, then \(n-1\) and \(n+1\) must be even. Now semiprimes can be even, if they are then \(2\) must be one of their prime factors. Because we are surmising \([n-1, n, n+1]\) are all distinct semiprimes, \(n-1\) and \(n+1\) must be even semiprimes.

Note that:

Given an even distinct semiprime, one factor must be \(2\) and the other (prime) factor must be odd.

^ The reason for this is because \(2\) is the only even prime number. Since we are considering only distinct semiprimes, the other prime factor for an even semiprime cannot be \(2\). Therefore, that prime factor must be odd. Suppose then that:

$$ \begin{aligned} n - 1 &= 2 * k \end{aligned} $$

where \(k\) is an odd number. The next even number, \([n - 1] + 2\), is expressed as:

$$ \begin{aligned} [n - 1] + 2 &= [2 * k] + 2 \cr &= 2\big([k] + 1\big) \cr &= 2\big(k + 1\big) \cr n + 1 &= 2\big(k + 1\big) \cr \end{aligned} $$

So! Now, we may now rewrite:

$$ \begin{aligned} n - 1, n, n + 1 \end{aligned} $$

as

$$ \begin{aligned} 2k, n, 2\big(k + 1\big) \end{aligned} $$

Given that if \(2k\) is semiprime, then \(k\) is odd, it is clear that \(k+1\) must be even. For this reason, \(n+1\) cannot be a semiprime.

From the analysis above, we arrive at two conclusions:

Given \(n\), an odd semiprime, either \(n-1\) or \(n+1\) is a semiprime but not both.

In otherwords, an odd semiprime cannot be preceeded and succeeded by a semiprime.

Also,

The longest possible semiprime chain when \(n\) is an odd semiprime is two.

A semiprime 🍞\(n\)🍞 where \(n\) is even.

If \(n\) is an even semiprime, then \(n-1\) and \(n+1\) are both odd. So, it is possible for one or both numbers (\(n-1\) and / or \(n+1\)) to be semiprime.

Quick intuition sidebar here| suppose we have two numbers \(a\) and \(b\) that are consecutively odd such that \(a < b \) (in other words, \(b = a + 2\)). Suppose now that \(b\) ends in \(5\). Then it is possible that \(b\) is a distinct semiprime (\(35, 55, 65\) are just a few examples of this). Also, if \(b\) ends in \(5\), then \(a\) must end in \(3\) as per our defintion that \(b = a + 2\). It is possible then for \(a\) to be a distinct semiprime if \(3\) is be a factor of \(a\) and the second factor of \(a\) is be any prime number > \(2\), such as \(11\) or \(31\). In fact, we have an example of this phenomenon from our primary example: if \(b == 35\), then \(a == 33\), both numbers are distince semiprimes.

Moreover, supposing \([n-1, n, n+1]\) do indeed form a semiprime sandwich (giving us a consecutive semiprime chain length of three), we know that \(n-2\) and \(n+2\) could not be distinct semiprimes because:

• \(n\) is an even distinct semiprime, so \(n = 2*k \) where \(k\) is odd (as per our reasoning in the previous section).
• \(n-2\) is also even but factors to \(n-2 = 2*(k-1)\).
  • So \(k-1\) must be even (since \(k\) is odd) and so it is not prime
  • pedantic note| (unless \(k-1 == 2\) but that would make \(n == 4\) which is a semiprime but not a distinct semiprime)
• likewise, \(n+2\) is also even but factors to \(n+2 = 2*(k+1)\).
  • So \(k+1\) must be even and therefore not prime

In short, we must conclude that:

IF \(n\) is an even distinct semiprime, then the longest possible chain of consecutive distinct semiprimes it may fall within must be \([n-1, n, n+1]\), with a length of 3.

Another interpretation of our analysis here is that a semiprime sandwich (consecutive distinct semiprimes of length 3) must have an even distinct semiprime as the middle component (or meat, if you will).

\(\big[33,34,35\big]\) is the first semiprime 🥪

The easiest way I could think of to show this was by first enumerating the first few even semiprimes. This is simple to do by hand, let’s take the first few primes:

$$ \begin{aligned} k &=[3,5,7,11,13,17] \cr 2*k &= [6,10,14,22,26,34] \cr \end{aligned} $$

(We multiply by \(2\) because we know that a semiprime sandwich must have an even semiprime as the middle element)

For each of the results above, look before and after and analyze if distinct semiprime or not, eg:

$$ \begin{aligned} k &= 6 \cr k - 1 &= 5 \cr k + 1 &= 7 \cr \end{aligned} $$

Since \(5\) is a prime number, we know that \(\big[5, 6, 7\big]\) is not a valid semiprime sandwich. For this analysis, \(34\) (and \(33\), \(35\)) emerge as the first example of a semiprime sandwich.

PS!

A random side note here: an “almost prime” or k-prime number is:

A natural number with k prime factors. For example, semiprimes can be thought of as “almost prime” with k=2. To that end, a prime number is a k-prime where k=1

I don’t know how to prove this yet (and I could be very wrong here!) but I would bet that the longest consecutive sequence of k-primes possible is always k+1. For instance, with primes, the longest consecutive sequence possible is 2, eg: \([2,3]\) since primes can be thought of as k-prime numbers where k=1. For semiprimes, eg: k-primes where \(k==2\), the longest consecutive sequence possible is 3, as shown here, etc.

UPDATE/PS: I proved myself wrong! Read more here

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