Many of us know the first few digits of \(\pi\) as \(3.14\). Some of us know \(\pi\) to higher accuracy — \(3.1415926\). And the few elite know \(\pi\) to thousands of digits. There are competitions for this!

Now, with probably zero practical math use, but 100% recreational math use, let’s see if we can find a sequence of prime numbers that contains \(\pi\). Here’s how this works.

Let’s consider the increasing precision sequence for the digits of \(\pi\) as \(p_{n}\) with the first few terms $$3,31,314,3141,31415,314159,3141592,31415926,\ldots$$

The task is to find the sequence of smallest prime numbers that contains \(p_{n}\) to various places. Thus, the smallest prime number that contains \(p_{0}\) is \(3\), which we will call \(a_{0}\). From here, \(a_{1}\) is the smallest prime number that contains \(p_{1}\) — i.e., \(a_{1} = 31\). And, \(a_{2} = 13147\) since \(314\) is in \(13147\). We continue like this for all \(p_{n}\).

The first several terms of \(a_{n}\) are \(\{3,31,13147,73141,314159,314159,\ldots\}\). Notice that \(314159\) shows up twice since the smallest prime number that contains both \(31415\) (\(p_{4}\)) and \(314159\) (\(p_{5}\)) is \(314159\).

The challenge is to find the next few terms! I don’t know of a good way to do this other than to program a search. Is this worthy of you, OEIS?? [Edit 7/25: Indeed! This sequence got published as A283247. Hurray!]

Good luck!