such that (in other words is an idempotent element of the 10-adic numbers). Of course, with a little algebra we can turn this into . In other words, both and are zero divisors in the set of 10-adic numbers.
In the post Foxmaths discusses how to generate other numbers of this form. In particular, assume is such a number and is the integer which corresponds to the first digits of . Foxmaths goes on to write:
The n-th digit of is given by the n-th digit of . Which I think is a strikingly simple result.
Observe. Starting at , we have . So the 1-th digit of is given by the 1-th digit of 25, the 2. Thus .
Carrying on, we see that , so the 2-th digit of is given by the 2-th digit of 625, the 6. .
Next, . So the 3-th digit of is a 0. This means that in value.
But then we have again, , and the 4-th digit of is given by the 4-th digit of 390625, the 9. Thus, .
And so it goes. We can use this to deterministically calculate as many digits of F as we like.
If I am reading his derivation correctly, then it appears the right-most digit of determines the rest of the digits of .
In particular, there must only be ten idempotent 10-adic integers (two of which are 0 and 1). [Update: Moreover, this digit can only be 0, 1, 5, or 6 because we need (mod 10). In particular, there must be only four idempotent 10-adic integers (two of which are 0 and 1).] Very cool.
Disclaimer: I do not know much more about p-adic numbers than what I’ve written on my blog. Presumably this fact is well-known among experts, or maybe I’m misinterpreting this and I’m completely wrong.