Are there other accumulation functions that holds $a(n-t)={a(n) \over a(t)}$? This might be a beginner's question regarding accumulation methods and their functions, but so far I have learned that compound interest satisfy 
$$a(n-t)={a(n) \over a(t)}$$
Which allows nice results such as 
$$s_{\bar{n}\rceil} = (1+i)^n a_{\bar{n}\rceil}$$
I also understand that the first property does not hold for simple interest.  But out of curiosity, are there accumulation functions that are not compound interests that holds the same property?  If so, does it show up in real life actuarial problems?
 A: You can rearrange your equation by multiplying through to get $a(n-t) a(t) = a(n)$.  Let $n = s+t$ and you get $a(s) a(t) = a(s+t)$.   So you want to find a function satisfying $a(s) a(t) = a(s+t)$ for all choices of $s$ and $t$.
Now, you can rewrite this to get $a(s) a(s) = a(2s)$, and then $a(s) a(2s) = a(3s)$ so $a(3s) = a(s)^3$, and in general $a(ms) = a(s)^m$ for any real number $s$ and integer $m$.  If you let $s = 1/m$, then you have $a(1) = a(1/m)^m$ and so $a(1/m) = a(1)^{1/m}$.  Therefore, for any integers $p$ and $q$, you have $a(p/q) = a(1/q)^p = (a(1)^{1/q})^p = a(1)^{p/q}$.  Thus if we know $a(1)$, we know $a(p/q)$ for any rational number $p/q$.  If you assume that the function $a$ is continuous then you can prove that any function satisfying your initial rule satisifes $a(x) = a(1)^x$ -- which, to go back to your initial question, is just the rule for compound interest.
A: The only real rule of accumulation functions is that a(0)=1.  So any function that you can think of for which f(n-t) = f(n)/f(t) qualifies provided you define a(0)=1.  An easy one would be a(t)=e^k(t)^2.  I'm just a student so I don't know what comes up in practice though.  
