There’s not much of a story to this post, except for a few curiosities the decimal system throws up (largely as a result of the binomial expansion).

Some time ago, I looked at some Fibonacci witchcraft: $\frac{1}{999,998,999,999} = 0.000\,000\, 000\,001\, 000\,001\, 000\,002\, 000\,003\, 000\,005\, 000\,008\,…$, neatly enumerating the Fibonacci sequence in six-digit blocks (which, like all of the following, can be lengthened by judiciously adding 9s to the denominator).

I came across some other neat ones. There’s the power series:

$\frac{1}{999} = 0.001\,001\,001\,…$ $\frac{1}{998} = 0.001\,002\,004\,…$ $\frac{1}{997} = 0.001\,003\,009\,…$

These are based on $(1-x)^{-1} = 1 + x + x^2 + x^3 + …$, for small $x$.

There’s also the Pascal’s triangle series:

$\frac{1}{999} = 0.001\,001\,001\,…$ $\frac{1}{999^2} = 0.001\,002\,003\,…$ $\frac{1}{999^3} = 0.001\,003\,006\,…$

My favourite, though, is this monster:

$\frac{1\,001\,000}{999^3} = 0.001\,004\,009\,016\,…$, giving the square numbers! (This is the same as $\frac{10^6}{999^3} + \frac{10^3}{999^3}$, which adds the third diagonal of Pascal’s triangle (the triangular numbers) to itself, offset by one – making the triangles into squares!

Someone with enough patience ought to be able to generate the cubes or higher powers with a bit of work.