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.

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

I'd tell you a dad math joke, but the good ones never quite reach me, and I don't want to make an asymptote of myself.

— Gregory Taylor (@mathtans) June 13, 2019

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

And we have this from MichLampinen

Q: How long is a mile?

A: Iβm not exactly sure, but I do know that itβs father than a kilometer!

— πΌπππππππ π»πππππππ (@MichLampinen) June 15, 2019

### 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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