OU blog

Personal Blogs

Richard Walker

A Very Surprising Number Pattern (And A Neat Proof It Goes On For Ever)

Visible to anyone in the world
Edited by Richard Walker, Tuesday 1 September 2026 at 00:01

sketch%20%284%29.png

Isn't that neat? It was posted on Quora by Alexander Ferrugia, along with a proof that the pattern goes on forever. His proof was quite long though, and I found one which is a bit shorter.

Something that jumped out at me was the resemblance to recurring decimals. one divided by six equals 0.1666 times ellipsis , one divided by two equals 0.5000 times ellipsis and one divided by three equals 0.3333 times ellipsis and I thought there had to be a connection. So I took the second line (the first doesn't fully show what the pattern is, because 6 doesn't appear) and played with it in the Python Shell, trying to get 16, 50 and 33, the numbers on the left hand side ...

100/6, 100/2, 100/3
(16.666666666666668, 50.0, 33.333333333333336)

Of course that's not right, we only want the whole number part. So let's subtract the fractional parts...

100/6 - 2/3, 100/2, 100/3 - 1/3
(16.0, 50.0, 33.0)

... and now it works. What about the right hand side? If we break it up we see it represents 16 multiplication 10000 plus 50 multiplication 100 plus 33 equals left parenthesis left parenthesis 10 squared solidus six minus two solidus three right parenthesis multiplication 10 squared multiplication 10 squared plus left parenthesis 10 squared solidus two right parenthesis multiplication 10 squared plus left parenthesis 10 squared solidus three minus one solidus three right parenthesis .

To get the equivalent expressions for the n -th line we can just replace 10 squared by 10 super n everywhere and now we get 

left parenthesis 10 super n solidus six minus two solidus three right parenthesis multiplication 10 super n multiplication 10 super n plus left parenthesis 10 super n solidus two right parenthesis multiplication 10 super n plus left parenthesis 10 super n solidus three minus one solidus three right parenthesis , with the left-hand side becoming sum with 3 summands left parenthesis 10 super n solidus six minus two solidus three right parenthesis cubed plus left parenthesis 10 super n solidus two right parenthesis cubed plus left parenthesis 10 super n solidus three minus one solidus three right parenthesis cubed

We want to prove these two expressions are equal and we can do that very easily. First make them more general by replacing 10 super n by x , giving 

left parenthesis x solidus six minus two solidus three right parenthesis multiplication x multiplication x plus left parenthesis x solidus two right parenthesis multiplication x plus left parenthesis x solidus three minus one solidus three right parenthesis and sum with 3 summands left parenthesis x solidus six minus two solidus three right parenthesis cubed plus left parenthesis x solidus two right parenthesis cubed plus left parenthesis x solidus three minus one solidus three right parenthesis cubed

It is not hard to see that if these were expanded we would obtain cubic polynomials of the form sum with 4 summands a times x cubed plus b times x squared plus c times x plus d . How can we prove they are the same? We might use paper and pencil to expand them and collect terms in the good old fashioned way. Or we could enter them into reliable computer algebra system and let it do the heavy lifting.

Or much more elegantly we can use this fact

If two cubic polynomials of the form sum with 4 summands a times x cubed plus b times x squared plus c times x plus d take identical values for four different values of x they are equal.

Why is this true? Because four values let us set up four equations and solve them to find the four constants a comma b comma c comma d .

But we have the requisite four values! This post began with them and so we have proved the two expressions are equal for any value of x and for any value of 10 super n in particular.

Permalink
Share post