name of mathematical progression

messing around on my calculator today, I discovered a fascinating sequence:


I’m sure somebody noticed this before, and had this named after him.
what’s it called?

I’m not sure they have a name (or need one).

The formula for the nth triangular number is T[sub]n[/sub] = ½n(n+1), from this it is easy to show that T[sub]n[/sub][sup]2[/sup] - T[sub]n-1[/sub][sup]2[/sup] = n[sup]3[/sup]

Probably not what you were hoping for, but A000537.

Nicomachus’s theorem. It is also called Aryabhata’s Identity, although Aryabhata lived four centuries after Nicomachus.

So, adding up triangular cubes equals triangular squares HAS NO special name?


Thanks for the OP. I had not noticed that. Working out why it holds gave me a chance to exercise my long unused proof by induction skills.

Even better, notice that the squares on the right side are the sum of the first n numbers. E.g.
1^3 + 2^3 + 3^3 + 4^3 + 5^3 = (1 + 2 + 3 +4 +5)^2

But I never knew any name for this.

That’s what “triangular numbers” means.


Here’s the simplest proof I know, though others will surely feel differently:

Consider 1[sup]p[/sup] + … + n[sup]p[/sup] for integer n. Note that this is always given by a (p+1)-th degree polynomial in n with leading coefficient 1/(p + 1), sending 0 to 0 and 1 to 1; furthermore, this polynomial will be symmetric around -0.5 for odd p and antisymmetric around -0.5 for nonzero even p.

Thus, T[sub]3/sub and (T[sub]2/sub)[sup]2[/sup] are both given by 4th degree polynomials with leading coefficient 1/4, sending 0 to 0 and 1 to 1, symmetric around -0.5. This leaves no degrees of freedom for them to differ, and thus they are the same.

Whoops, I made a mistake in editing: I meant to say “Let T[sub]p/sub be 1[sup]p[/sup] + … + n[sup]p[/sup] for integer n”, so I could refer to that name again later.

Also, I think you meant:
Thus, T[sub]3/sub and (T[sub]1/sub)[sup]2[/sup] are both given by 4th degree polynomials …


Yes, good catch!

Yes, but does .999 = 1?

Sorry, couldn’t resist.