Let’s get to the heart of it right away, and pick a totally non-arbitrary number, 7, just because it’s my favorite one-digit number. The reciprocal (or multiplicative inverse) of 7 can be written in decimals as the following shows:

The bar on top of the 142857, also called a vinculum, marks these digits as being recurring (see repeating decimals), meaning that

an that, to infinity (or for eternity, if you actually try to calculate it).

Do you see something strange about 142857? It starts with 14, then adds 28 (14×2), then it almost has 56 (14×4). But try to calculate the next number, and you get 112 (14×8). If you add the first few numbers, you get the following:

which is summed as

As you do more and more iterations, more digits are fixed and the result gets closer and closer to the expected value (it *converges* towards that value). For example, the next iteration (…3584) will fix the 39 at the end of the sum to the correct 42. And so on and so forth.

Are there any other numbers that have the same property? After all, this seems like a pretty weird coincidence.

Actually, *all* numbers’ reciprocals behave as such. Hn hnnn, it’s true! More to follow.

Are there any other sum of numbers that will yield the recurring 142857? To be honest, the previous addition seems like luck, or like there’s something alien about it.

Actually, it would seem alien to anyone who has not studied number theory, I guess. There are an infinity of such sums that give you a recurring 142857. I need to repeat that: *an infinity of such sums exist*.

To make the rest of the post easier to read, let’s define these sums. In the previous example, you begin with 14, and then multiply it repeatedly by 2, moving your position (precision) by two each time. We can “encrypt” that precision (position moving on each iteration) in the starting value: 14 means you move by two, and 014 means you move by three. I could then write the above sum as **14××2** (I couldn’t find a suitable, untaken character for this).

So, what other sums can give you 142857 for infinity? How about 142××6? Try it out! Then try 1428××4, then 14285××5, and 142857××1. Also, don’t forget to check 1××3, and 1428571××3. Right there, you can see that you can have an infinity of possible sums.

Although they’re based on the same numbers, it’s still interesting, right? No, it gets better. A lot. Try 13××9. Honestly. Pick a piece of paper and try it. Here, let me help you with it:

Granted, it converges slowly, but you can easily see that, given some more iterations, you get the expected value. (In fact, the farther you are from ××1, the slower it converges.)

Now, try 12××16, and 11××23. You can already see a trend, right? Down to 01××93.

Let me also add 2××-4, and 15××-5. Up to 28××-96. Then I could pick random ones, like 087××391, or 1431××-17.

**Yes, all of them converges to a recurring 142857.**

If you list them all, it’s pretty easy to find a general formula:

where k is the precision, v is the arbitrary integer number you want to start with, and 0 < v < 2×10^k/7. It’s that easy! For example, if you want a number with three digits of precision (*k*), you can take any initial value *v* that is between 1 and 285 (2×1000/7). *v* = 0 results in a division by zero, and other values make the series diverge. Actually, you can do the same with negative values of *v*, except that the result is negative.

Remember when I talked about numbers different than 7 having the same property? You probably noticed (or wondered) if the 7 in the general formula above can be replaced by *i*, which would be the source of the reciprocal. Let’s define the new general equation more formally:

And what’s incredible with that, is that every reciprocal can be calculated that way. You have your *i* value at the start (the reciprocal you want), then arbitrarily choose any precision (*k*) and multiplier (*v*), and you have an equation that matches a set of recursive numbers to infinity.