When thinking about binary representations of triangular numbers, I noticed an interesting property:

In the cases I've tested, for the numbers from $0$ to $2^n-1$, each combination of the last $n$ bits occurs exactly once, that is, $k\mapsto k(k+1)/2 \bmod 2^n$ is a bijection on the set $\{0,\ldots,2^n-1\}$.

Or stated differently: For those $n$ I tested, all numbers are triangular modulo $2^n$.

That rises two related questions:

1. Does this hold for every $n$?
2. What happens modulo a number $N$ that's not of the form $N=2^n$?

Or short: For which $N$ are all numbers triangular modulo $N$?

Now it is easily checked that this cannot hold for odd $N$ other than $N=1$, since in that case $(N-1)N/2 \equiv 0 \pmod N$ because the denominator does not cancel out any factor in $N$.

I've checked with Python code for $N<10000$, and found that for those, it's exactly the powers of $2$ that fulfil the condition.

Therefore my conjecture is:

All numbers are triangular modulo $N$ iff $N$ is a power of $2$.

However I have no idea how I could proof (or disproof, other than by a counterexample, which I've obviously not found) this conjecture.

Can you shed some light on it?

posted 7 months ago

CC BY-SA 4.0

7mo ago

celtschk‭

1011 reputation 17 26 135 29