<s>[INST] Prime sum identity Let $ \Lambda(k) $ denote the von Mangoldt function:
$$
\Lambda(k) \stackrel{\text{def}}{=}
\begin{cases}
0      & \text{if $ k $ is not a prime power}, \\
\ln(p) & \text{if $ k = p^{j} $}.
\end{cases}
$$
Also, let $ \lfloor x \rfloor $ be the floor function.
Can anyone prove any of the following identities:
$$
\sum_{k=0}^{\left\lfloor (n - 1)/3 \right\rfloor} \ln(3k+1) = \sum_{k=0}^{\left\lfloor (n - 1)/3 \right\rfloor} \Lambda(3k + 1) \left\lfloor \frac{n}{9k + 3} + \frac{2}{3} \right\rfloor + \sum_{k=0}^{\left\lfloor (n - 2)/3 \right\rfloor} \Lambda(3k + 2) \left\lfloor \frac{n}{9k + 6} + \frac{1}{3} \right\rfloor.
$$
$$\sum_{k=0}^{\left\lfloor (n - 3)/4 \right\rfloor} \ln(4k+1)=\sum_{k=0}^{\left\lfloor (n - 1)/4 \right\rfloor} \Lambda(4k + 1) \left\lfloor \frac{n}{16k + 4} + \frac{1}{4} \right\rfloor + \sum_{k=0}^{\left\lfloor (n - 3)/4 \right\rfloor} \Lambda(4k + 3) \left\lfloor \frac{n}{16k +12} + \frac{3}{4} \right\rfloor.$$
$$\sum_{k=1}^{n} \ln(k)= \sum_{k=1}^{n} \Lambda(k) \left\lfloor \frac{n}{k} \right\rfloor$$
The latter ones obvious, I just thought id post it for comparison.
 [/INST] A perhaps simpler proof, using the $p$-adic expansion of $n!$:
$$\log(n!)=\sum_{p \text{ prime}}\log(p)\left(\left \lfloor \frac{n}{p} \right \rfloor+\left \lfloor \frac{n}{p^2}  \right \rfloor+...\right)=\sum_{k=1}^n \Lambda(k)\left \lfloor \frac{n}{k}  \right \rfloor. $$
</s>