# Last digits of iterated triangular numbers

First, let me explain what I mean by iterated triangle numbers:

The triangle of a number $n$ is the sum of all numbers from 1 to n, i. e. $1+2+...+ n$. Iterated triangle numbers are what you get if you repeat this process (that is, find the triangle of a number, take the result, and then triangle again).

Now, on to my question. If you repeat this process over and over again, do the last digits ever repeat? If you take 2 and find the triangle repeatedly, you get 3, 6, 21, 231, 26796, 359026206, 64449908476890321, .... The next few numbers in the series have the last digits ...186681, ...991221, and ...531031. The 100th iterated triangle of 2 also ends in ...453031 (yes I actually took the time to find that out).

The last digit is always either 1 or 6, but which one it is is quite erratic, and I know of no way to find the last digit of, say, 2 triangled 443 times, without finding the last digits of all numbers in the series up to it.

• The earlier answer was deleted without comment so I'd like to point out what was wrong with it: the assertion that $n \equiv a \implies T(n) \equiv T(a) \pmod{10^k}$ fails because $n(n+1)/2$ does not have integer coefficients. It doesn't appear that the sequence is known to be periodic, but at least it has been tabulated at oeis.org/A117872. (That is to say, the parity is enough to distinguish between the last digits $1$ and $6$.) – Erick Wong Feb 22 '18 at 11:20