# Guest post: Sequence Numbers

This is a guest post, sent in by David, who’s discovered an interesting property of numbers, and is looking for collaborators to take it further.

Mathematics is not a careful march down a well-cleared highway, but a journey into a strange wilderness, where the explorers often get lost. Rigour should be a signal to the historian that the maps have been made, and the real explorers have gone elsewhere.

– W. S. Anglin

A few years ago I saw a post on a website that showed that the inverse of 998,001 produces a decimal expansion that counts, using three digit strings, from 000 to 997 without error.

$\frac{1}{998,\!001} = 0.0000010020030040050060070080090100110120130\ldots$

I immediately thought that this had to be a hoax. I decided to work it out to prove it was a hoax – after all some people put anything they want on the web whether it is true or not.

My calculator cannot handle more than eight digits. And I really did not want to work it out by hand. The arithmetic is simple enough, I just did not want to spend that much time on it (we all like instant everything – personally I get impatient waiting on the microwave). So I took this problem to a wonderful website, Wolfram|Alpha. It did the calculation for me, and I checked it out – digit by digit. I did not like what I found out – the equation was right, and I was wrong.

But I did start wondering – I wondered if there were other numbers that had similar properties as this one. Are there other integers that, when you turn the number into a unit fraction and then convert it to a decimal, produce an interesting sequence of numbers (like counting, or multiplying, or whatever).

So I looked on the web for answers – but the internet did not have any answers to this question (stupid internet!). Could it be true that the example I found in a blog post was the only example of this special property?

I contacted a few other mathematicians by e-mail to see if they knew anything about this kind of a mathematics – most did not – but a few gave me so hints on where I might find out more. But those websites only gave me more hints.

By this time I realized two things – I was not going to find an answer out there – and if I wanted to know the answer I was going to have to find it myself. And I really wanted to know the answer!

Things were slow at first, and at second, and at third… but I found a few more hints, and eventually found another number with a similar properties.

Then I found a few more.

But I did not understand why some numbers worked and others did not. So I kept working on the problem, and as I found more of these special numbers I started to see patterns, and I learned how to fill in some blank spots in these patterns.

I decided this group of numbers needed a special name – after all I could not keep calling them just “special numbers. “Mr. B’s Super-duper-fantastic Arithmetically Accurate and Mathematically Mystical Sequence Numbers”. This name is really kind of long and it did not catch on. Now I just call them “Sequence Numbers” because they produce interesting sequences.

I define sequence numbers as integers that produce interesting sequences when you take their inverses and convert them into decimal form. I prefer that the “interesting sequence” is listed in the Online Encyclopedia of Integer Sequences, but if the sequence is an obvious or well known sequence I can accept that too.

So what are some of these other numbers? Let me tell you about some of my new buddies:

• 9,899 produces the first few terms of the well known Fibonacci sequence.
• 196,020 produces a list of the multiples of five.
• 998,999 produces the terms of the Fibonacci sequence, using three digit strings.
$\frac{1}{998999} = 0.000 \; 001 \; 002 \; 003 \; 005 \; 008 \; 013 \; 021\ldots$
• 998,998 produces the terms of the Jacobsthal sequence, using three digit strings.
$\frac{1}{998998} = 0.000 \; 001 \; 001 \; 003 \; 005 \; 011 \; 021 \; 043 \; 085 \; 171\ldots$
• 76,922,923,077 counts by multiples of 13. (Imagine that! How far can you count by multiples of 13?)
• 999,999,700,000,029,999,999 shows a list of all of the triangular numbers with 6 digits or less, and even most of the triangular numbers with seven
• 999,999,999,997 lists all of the powers of two ($2^0$, $2^1$, $2^2$, $2^3$, etc.) up to 12 digits long, except for the largest 12-digit power of 2. And it writes them all in 12-digit strings.
• 999,999,998,999,999,998,999,999,998,999,999,998,999,999,998, 999,999,998,999,999,998,999,999,998,999,999,998,999,999,998, 999,999,998,999,999,998,999,999,999 shows the terms of the tridecanacci sequence, defined as $a(n) = \begin{cases} 0 & 0 \leq n \lt 12, \\ 1 & n=12, \\ \sum_{i=1}^{13}a(n-i) & n \gt 12 \end{cases}$
• 999,999,999,876,543,210 – this one shows the 123,456,789 times table (all of the multiples of 123,456,789) up to, but not including the last 24 digit multiple of 123,456,789.

There are a lot of these special numbers out there, and I know I have not found them all.

I hope some of you don’t believe me. Well let me rephrase that – I hope that some of you don’t believe me AND have the gumption to try to prove me wrong. If you do you will learn how to find these numbers, and even how to design some to do specific things.

If you find something you want to show me, or you want me to post for others, please email me at: mbiom.edu@gmail.com. Check out my website at: SequenceNumbers.blogspot.com.

Don’t be afraid of the big numbers. That may be where all the really interesting and fun stuff starts.

### 4 Responses to “Guest post: Sequence Numbers”

1. Thomas Oléron Evans

These are rather neat.

The fraction that lists powers of 2 can be derived using the formula for the sum of a geometric series (though the one you’ve listed actually gives powers of 3, I think; 1/(10^k – n), gives the powers of n up to k digits, in general).

Also, Colin Beveridge has a clear explanation of how 1/999999999998999999999999 generates the Fibonacci sequence here:

2. Stephen Miller

The original example generalises nicely. Say we want to list (almost) all k digit numbers; then we first need to know that
$\frac{1}{10^k-1}=0.\dot 0…\dot 0 \dot 1$
where the recurring term is $k-1$ zeroes followed by a one. Then we need to add a few of those up – well, I won’t spoil ALL the fun, but we get the nice sequence number
$\frac{1}{(10^k-1)^2}$

$f(x) = \frac{x^2}{(1-x)^2} = 0x + 1x^2 + 2x^3 + 3x^4 + \dots \,.$
Then it is $f(\frac{1}{1000}) = \frac{1}{998001}$. So take the generating functions of your favourite number and just plug in $10^{-d}$ to get a number which has the sequence at their decimalplaces. The $d$ just gives the number of “space” they have.