Help with counting factors Can someone help me prove that the product below has $(\large\frac{10^{100}}{6}-\frac{17}{3})$ factors of $7$?
$$\large\prod_{{k}={1}}^{{{10^{100}}}}({4k}+1)$$
 A: Below is a partial answer.
Let $S_n = \displaystyle \prod_{k=1}^n(4k+1)$.
Number of multiples of $7$ in $S_n$ is $\left \lfloor \dfrac{n +2}7 \right\rfloor$
Number of multiples of $7^2$ in $S_n$ is $\left \lfloor \dfrac{n +37}{49} \right\rfloor$
Number of multiples of $7^3$ in $S_n$ is $\left \lfloor \dfrac{n +86}{343} \right\rfloor$
Number of multiples of $7^{2k}$ in $S_n$ is $$\left \lfloor \dfrac{n + 7^{2k} - (7^{2k}-1)/4}{7^{2k}} \right\rfloor = \left \lfloor \dfrac{4n + 3\cdot 7^{2k} + 1}{4 \cdot 7^{2k}} \right\rfloor$$
Number of multiples of $7^{2k+1}$ in $S_n$ is $$\left \lfloor \dfrac{n + 7^{2k+1} - (3 \cdot 7^{2k+1}-1)/4}{7^{2k+1}} \right\rfloor = \left \lfloor \dfrac{4n + 7^{2k+1}+1}{4 \cdot 7^{2k+1}} \right\rfloor$$
Hence, the highest power of $7$ dividing $S_n$ is
$$\sum_{k=1}^{\infty} \left(\left \lfloor \dfrac{4n + 7^{2k-1}+1}{4 \cdot 7^{2k-1}} \right\rfloor + \left \lfloor \dfrac{4n + 3\cdot 7^{2k} + 1}{4 \cdot 7^{2k}} \right\rfloor \right)$$
I computed the sum above using WolframAlpha and it agrees with $\dfrac{10^{100}}6 - \dfrac{17}3$
