I have an algorithm that is based on a tree and depends on two input sizes n and m, where n is the number of edges a node can have and m is the number of leaves.

I want to find Big-O of an algorithm that traverses said tree and it breaks down to the following formula for the number of nodes.

$m = n^x$

$x = \log_n(m)$

$$\frac{n^{x+1}-1}{n - 1} = \frac{n\cdot n^x-1}{n - 1} = \frac{n\cdot n^{\log_n(m)}-1}{n - 1} = \frac{nm - 1}{n-1} = O(?)$$

Is Big-O of the Formula $O(mn)$ or $O(m)$ because $n$ cancels out? Thank you in advance!

PS: I know that if $O(m)$ is the upper bound $O(mn)$ is an upper bound too, I want to find the tightest.

I have an algorithm that depends on two input sizes n and m. The complexity breaks down to the following equation:

$\frac{nm - 1}{n-1} = O(?)$

Is Big-O of the Formula $O(mn)$ or $O(m)$ because $n$ cancels out? Thank you in advance!

PS: I know that if $O(m)$ is the upper bound $O(mn)$ is an upper bound too, I want to find the tightest.

I have an algorithm that is based on a tree and depends on two input sizes n and m, where n is the number of edges a node can have and m is the number of leaves.

I want to find Big-O of an algorithm that traverses said tree and it breaks down to the following formula for the number of nodes.

$m = n^x$

$x = log_n(m)$

$\frac{n^{x+1}-1}{n - 1} = \frac{n*n^x-1}{n - 1} = \frac{n*n^{log_n(m)}-1}{n - 1} = \frac{n*m - 1}{n-1} = O(?)$

Is Big-O of the Formula $O(m*n)$ or $O(m)$ because $n$ cancels out? Thank you in advance!

PS: I know that if $O(m)$ is the upper bound $O(m*n)$ is an upper bound too, I want to find the tightest.

I have an algorithm that is based on a tree and depends on two input sizes n and m, where n is the number of edges a node can have and m is the number of leaves.

I want to find Big-O of an algorithm that traverses said tree and it breaks down to the following formula for the number of nodes.

$m = n^x$

$x = \log_n(m)$

$$\frac{n^{x+1}-1}{n - 1} = \frac{n\cdot n^x-1}{n - 1} = \frac{n\cdot n^{\log_n(m)}-1}{n - 1} = \frac{nm - 1}{n-1} = O(?)$$

Is Big-O of the Formula $O(mn)$ or $O(m)$ because $n$ cancels out? Thank you in advance!

PS: I know that if $O(m)$ is the upper bound $O(mn)$ is an upper bound too, I want to find the tightest.