Alternative way to calculate number of edges in Turán-graphs? We define a Turán-graph
$$T_n(r) = K_{\lceil\frac{n}{r}\rceil, \ldots,\lceil\frac{n}{r}\rceil, \lfloor\frac{n}{r}\rfloor, \ldots, \lfloor\frac{n}{r}\rfloor}$$
with $n \bmod r$ many subsets $V_i$ that contain $\lceil\frac{n}{r}\rceil$ many vertices and $r- (n \bmod r)$ many subsets with $\lfloor\frac{n}{r}\rfloor$ vertices. Let $n=pr+q$. We calculate the number of edges in such a graph as follows $$t_n(r) = \left( 1 - \frac{1}{r}\right)\frac{n^2}{2} - \frac{q(r-q)}{2r}$$
My thoughts:
We can think of Turán-graphs as $r$-partit graphs. So for some $T_n(r) = K_{V_1, V_2, \ldots, V_r}$ with $$e_{i} = \binom{i}{2}$$ representing the number of edges in some complete graph $K_i$.
Hence, we can calculate the number of edges as follows:
\begin{align*}
t_n'(r) &= e_{n} - e_{V_1} - e_{V_2} - \ldots - e_{V_r} \\
&= \binom{n}{2} - \binom{V_1}{2} - \binom{V_2}{2} - \ldots -\binom{V_r}{2}\end{align*}
Question:
This seems way more simple to me than the formula given above. If $t_n'(r)$ works correct (couldn't prove it wrong nor right), why bother using $t_n(r)$?
 A: Your formula is fine. It is essentially what Wikipedia's article is using in the introduction at the moment, for example. I think it's not as simple as you believe: to have a complete formula we should include the sizes of the parts, which gives us $$t'(n,r) = \binom n2 - (n \bmod r) \binom{\lceil n/r\rceil}{2} - (r - (n \bmod r))\binom{\lfloor n/r \rfloor}{2}.$$
It is still a perfectly good formula.
The advantage of $t(n,r)$ is that it tells us the "nice approximation" $(1 - \frac1r) \frac{n^2}{2}$ first, and then it gives the error in that approximation.
Watch out for the incorrect formula
$$
    \left\lfloor \frac{(r-1)n^2}{2r}\right\rfloor = \left\lfloor \Big(1 - \frac1r\Big)\frac{n^2}{2}\right\rfloor
$$
found on Wolfram MathWorld and other sources that cite it. It agrees with the correct formula in many cases, in particular for all small values of $r$. This formula is first wrong when $n=12$ and $r=8$: it gives $63$, while the number of edges in $K_{2,2,2,2,1,1,1,1}$ is only $62$. In general, it may be too high by as much as $\frac r8$: the upper bound on the $\frac{q(r-q)}{2r}$ error term in the first formula, where $q = n \bmod r$.
