A Loopy Sequence For Father’s Day

Happy Father’s Day!

Today’s the day that dads get ties and get to tell jokes with impunity. Or if you’re a mathematician, you get a book on knots so you can tie the perfect tie.

Now, I was able to get a Trinity tie put together.

Trinity Tieknot!

So my job now is to learn how to tie the Eldredge. It feels like I’m going to need a necktie that’s about 7 feet long to make this work. But hot damn.

Source: https://i.imgur.com/8ug05FA.jpg

My wonderful wife got me The 85 Ways to Tie a Tie: The Science and Aesthetics of Tie Knots

Also for today, I’ll put forth a “Father’s Day Sequence”. Keep reading to find out more!

First, let’s have some shoutouts from the Twitterverse.

We have this unimpeachable gem from @mathtans

Make sure to check out his website at https://mathtans.blogspot.com/.

And we have this from MichLampinen

Now getting back to ties

Ties are knots. Knots have loops. Let’s talk about a loopy sequence!

Let \(0,4,6,8,9\) be “loopy digits” (can you guess why?) and \(1,2,3,5,7\) be “stranded digits”.

Now, let’s make up our loopy sequence — an increasing sequence of non-negative integers with a special construction rule. First, some preliminaries. We will keep track of the number of “loops” for any given term in our loopy sequence. For the loopy digits, 0 has one loop, 4 has one loop, 6 has one loop, 8 has two loops, and 9 has one loop. The stranded digits have zero loops. A number like 63891 has four loops.

Let’s see how we construct our loopy sequence.

Our loopy sequence begins at 0. Thus, \(a_{0} = 0\) and if \(n_{k}\) is the number of loops for the \(k\)th term of the sequence, then \(n_{0} = 1\). The next number in the sequence, \(a_{1}\) is the first integer greater than \(a_{0}\) such that \(n_{1} \geq n_{0}\). Thus, \(a_{1} = 4\). We should be able to see that \(a_{2} = 6\). Do you believe that we must have \(a_{3} = 8\)? Since \(n_{2} = 1\), we want the first integer greater than \(a_{2}\) such that \(n_{3} \geq n_{2}\). The first integer that does this is \(a_{3} = 8\). Take a few minutes to reason this out if it seems confusing. If after a few minutes it is still confusing, then keep reading, maybe more examples will clarify the situation.

Now, what is \(a_{4}\). It’s not 9 since \(n_{4} = 1\) and we are already at \(n_{3} = 2\). So what’s the next integer that has at least two loops? It must be \(a_{4} = 18\). Makes sense?

Here are the first few terms of our loopy sequence.
$$0, 4, 6, 8, 18, 28, 38, 40, 44, 46, 48, 68, 80, 84, 86, 88$$
and here is what the loop count sequence
$$1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4$$

Here are a few questions for you to work out.

Can you write down the first 20 terms of this sequence?

Will the digits 5, 7, and 9 ever show up?

Is there a closed form solution to generate the \(k\)th term (\(k > 0\)) of the sequence or must we know the previous term?

If we wanted to create a different loopy sequence, \(B\), with \(b_{0} > 0\), then prove or disprove that there exists \(k, j \in \mathbb{N}\) such \(a_{k + i} = b_{j + i}\) for all non-negative integer \(i\).

In other words, how we start the loopy sequence doesn’t matter. Every loopy sequence will eventually align with our original construction. For example, if \(b_{0} = 11\), then we have \(11, 12, 13, 14, 16, 18, 28, \ldots\) and we see that \(a_{4 + i} = b_{5 + i}\) for all \(i \geq 0\).

If you were given the number of loops, \(n\), would you be able to write down the first possible integer to satisfy that requirement assuming \(a_{0} = 0\)?

While You’re Here …

Interested in knowing what Primal Words are? How about Funny Numbers? Or Semi r-Primes?

Happy Father’s Day!

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