Closed form for a recurrence relation $(p^n-p^{n-1})p^{n(k-1)}+p^{n-1}a_{k-1}=a_{k}$

Question

So I'm a bit rusty with recurrence relations and I was wondering if someone could help me with this. I'm basically trying to calculate the number of elements of order $p^n$ in a specific group and I got the following recursive relation $$(p^n-p^{n-1})p^{n(k-1)}+p^{n-1}a_{k-1}=a_{k}$$ Note that $n$ is a fixed value in this case. I would really appreciate any help I can get trying to find a closed form for the above expression.

Answer 1

From $\frac{a_k}{p^{(n-1)k}} =\frac{a_{k-1}}{p^{(n-1)(k-1)}}+\frac{p^n-p^{n-1}}{p^{n-k}} $, let $b_k =\frac{a_k}{p^{(n-1)k}} $.

Then $b_k =b_{k-1}+cp^k $, where $c =\frac{p^n-p^{n-1}}{p^{n}} =1-\frac1{p} $.

Therefore $b_k-b_{k-1} =cp^k $. Summing from $k=1$ to $m$, $b_m-b_0 =c\sum_{k=1}^{m} p^k =(1-\frac1{p}) \frac{p^{m+1}-p}{p-1} $ so $b_m =a_0+ (1-\frac1{p})\frac{p^{m+1}-p}{p-1} $ and you can get $a_m =p^{m(n-1)}\left(a_0+(1-\frac1{p}) \frac{p^{m+1}-p}{p-1}\right) =p^{m(n-1)}\left(a_0+p^m-1\right) $.