There are $n$ sticks of length $\{1,2,3,\cdots,n\} $. how many incongruent triangles can be made using those sticks

Question

My Approach: Let the number of incongruent triangles possible with the sticks of length $\{1,2,\cdots,n\}$ is $S_n $

So now number of incongruent triangles possible with the sticks of length $\{1,2,\cdots,n,n+1\}=$ Number of incongruent triangles with the sticks of length $\{1,2,\cdots,n\}+$ Number of incongruent triangles with largest side $n+1 $ $\implies S_{n+1}=S_n+$ Number of incongruent triangles with largest side $n+1 $

Now we will find the number of possible triangles with largest side $n+1$

Now the smallest side has to be greater than 1. The sum of two sides other than largest side can be minimum $n+2$ and maximum $2n-1$

\begin{array}{c|c} \text{Smallest Side length}& \text{Number of choices to choose third side}\\ \hline 2 & 1\ [\text{Only } n]\\ 3 & 2\ [\text{Only } n,n-1]\\ 4 & 3\ [\text{Only } n,n-1,n-2]\\ \vdots & \vdots \quad \quad\quad\vdots \\ \Big\lfloor\dfrac{n}{2}\Big\rfloor & \Big\lfloor\dfrac{n}{2}\Big\rfloor-1\\ \hline \text{Total} & \dfrac{\Big\lfloor\dfrac{n}{2}\Big\rfloor\bigg(\Big\lfloor\dfrac{n}{2}\Big\rfloor-1\bigg)}{2} \end{array}

Case 1 ($n=2k$): Then $\Big\lfloor\dfrac{n}{2}\Big\rfloor=k$ Then after k we can choose any two lengths from the sticks $\{k+1,k+2,\cdots,n\}$. So then number of choices when smallest side is $k+1$ is $\dfrac{k (k-1)}{2}$

Hence $S_{n+1}=S_n+\dfrac{k (k-1)}{2}+\dfrac{k (k-1)}{2}=S_n+k (k-1)= S_n+\dfrac{n (n-2)}{4}$

Case 2 ($n=2k+1$): Then $\Big\lfloor\dfrac{n}{2}\Big\rfloor=k$. Here after k we can choose any two lengths from the sticks $\{k+1,k+2,\cdots,n\}$. So now the number of choices when smallest side is $k+1$ is $\dfrac{k (k+1)}{2}$

Hence $S_{n+1}=S_n+\dfrac{k (k-1)}{2}+\dfrac{k (k+1)}{2}=S_n+k^2 = S_n+\dfrac{(n-1)^2}{4}$

Now i got these two recurrence relation but how to solve that to a general formula?

Edit: I got the rest.

We got that $$\text{Number of incongruent triangles with largest side }n+1=\begin{cases} \dfrac{n (n-2)}{4}, & \text{when }n=\text{even}\\ \dfrac{(n-1)^2}{4}, & \text{when }n=\text{odd}\end{cases}$$we can write it in this way that $$\text{Number of incongruent triangles with largest side }n+1=\bigg\lceil \dfrac{n (n-2)}{4}\bigg\rceil$$

Now, \begin{array}{ccccc} S_{n+1}& - & S_n & = & \bigg\lceil \dfrac{n (n-2)}{4}\bigg\rceil \\ S_{n} & - & S_{n-1} & = & \bigg\lceil \dfrac{(n-1)(n-3)}{4}\bigg\rceil \\ S_{n-1} & - & S_{n-2} & = & \bigg\lceil \dfrac{(n-2)(n-4)}{4}\bigg\rceil \\ \vdots & & \vdots & & \vdots\\ S_4 & - & S_3 & =& \bigg\lceil \dfrac{3 (3-2)}{4}\bigg\rceil =1\\ \hline S_{n+1}&- & S_3 & = & \sum\limits_{i=3}^n \bigg\lceil \dfrac{i (i-2)}{4}\bigg\rceil\\ S_{n+1}& & & = &\sum\limits_{i=1}^n \bigg\lceil \dfrac{i (i-2)}{4}\bigg\rceil \end{array}