This post is based on work by Mark Ritchings; I know of no finer ((or, now you mention it, other)) maths tutor in Bury.

A few weeks ago, I pointed in the vague direction of a few decimal curiosities – fractions that spit out lovely patterns in their decimal expansions. Having found one that generated the squares, I asked the world at large to find fractions that generated cubes or higher powers.

Mark (being Mark) responded within hours, pointing out:

$C=\frac{334667000}{332001998667}$ for the cubes, and $Q=\frac{333337000003666667000000}{333331666669999996666668333333}$ for the fourth powers.

In the interests of clarity, I’d probably write them in not-quite-lowest form:

$C = \frac{1,004,001,000}{(1000-1)^4}$ $Q = \frac{1,000,011,000,011,000,001,000,000}{(1000-1)^5}$

His method involved two of my favourite tricks (division by matching coefficients and solving using matrices), but I’m not going to attempt to reproduce his work here!

I’ve come at it from a different angle, using a binomial expansion and the diagonals of Pascal’s triangle ((discovered, of course, long before Pascal’s time.))

Applying a bit of C4 gives you the following: $(1-x)^{-1} = 1 + x + x^2 + x^3 + …$, which I’ll write as $S_1 = 1$. $(1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + …$, so $S_2 = n$. $(1-x)^{-3} = 1 + 3x + 6x^3 + 10x^3 + …$, so $S_3 = \frac{1}{2}n(n+1) = \frac{1}{2}\left(n^2 + n\right)$

Since you can hopefully see the expansion pattern by now, I’ll just state that $S_4 = \frac{1}{6}n(n+1)(n+2) = \frac{1}{6}\left(n^3 + 3n^2 + 2n\right)$ and $S_5 = \frac{1}{24}\left(n^4 + 6n^3 + 11n^2 + 6n\right)$.

Why is that important? Aha! If I want to generate $n^3$, which I do, I note that $6S_4 = n^3 + 3n^2 + 2n$. I can remove the $3n^2$ by way of subtracting $6S_3$, so that I have $6S_4 - 6S_3 = n^3 - n$; if I add $S_2$, I have what I need!

$\frac{6}{(1-x)^4} - \frac{6}{(1-x)^3} + \frac{1}{(1-x)^2} = 1 + 8x + 27x^2 + …$

That can be tidied up as $\frac{6 - 6(1-x) + (1-x)^2}{(1-x)^4} = \frac{1 + 4x + x^2}{(1-x)^4}$, as Mark got.

In a similar way, $n^4 = 24S_5 -36S_4 + 14S_3 - S_2$, so my expression for the generating sequence is $\frac{24}{(1-x)^5} - \frac{36}{(1-x)^4} + \frac{14}{(1-x)^3} - \frac{1}{(1-x)^2}$, or $\frac {24-36(1-x)+14(1-x)^2-(1-x)^3}{(1-x)^5}$, or even $\frac{1 + 11x + 11x^2 + x^3}{(1-x)^5}$ – again, exactly as Mark found!

I’d love to know of a less algebra-intensive method of finding these functions, though!