Prime Factors of 39 Factorial

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Example of Factorial

The prime decomposition of $39!$ is given as:

$39! = 2^{35} \times 3^{18} \times 5^8 \times 7^5 \times 11^3 \times 13^3 \times 17^2 \times 19^2 \times 23 \times 29 \times 31 \times 37$


Proof

For each prime factor $p$ of $39!$, let $a_p$ be the integer such that:

$p^{a_p} \divides 39!$
$p^{a_p + 1} \nmid 39!$


Taking the prime factors in turn:

\(\ds a_2\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {39} {2^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {39} 2} + \floor {\frac {39} 4} + \floor {\frac {39} 8 } + \floor {\frac {39} {16} } + \floor {\frac {39} {32} }\)
\(\ds \) \(=\) \(\ds 19 + 9 + 4 + 2 + 1\)
\(\ds \) \(=\) \(\ds 35\)


\(\ds a_3\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {39} {3^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {39} 3} + \floor {\frac {39} 9} + \floor {\frac {39} {27} }\)
\(\ds \) \(=\) \(\ds 13 + 4 + 1\)
\(\ds \) \(=\) \(\ds 18\)


\(\ds a_5\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {39} {5^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {39} 5} + \floor {\frac {39} {25} }\)
\(\ds \) \(=\) \(\ds 7 + 1\)
\(\ds \) \(=\) \(\ds 8\)


\(\ds a_7\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {39} {7^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {39} 7}\)
\(\ds \) \(=\) \(\ds 5\)


\(\ds a_{11}\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {39} {11^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {39} {11} }\)
\(\ds \) \(=\) \(\ds 3\)


\(\ds a_{13}\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {39} {13^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {39} {13} }\)
\(\ds \) \(=\) \(\ds 3\)


\(\ds a_{17}\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {39} {17^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {39} {17} }\)
\(\ds \) \(=\) \(\ds 2\)


\(\ds a_{19}\) \(=\) \(\ds \sum_{k \mathop > 0} \floor {\frac {39} {19^k} }\) De Polignac's Formula
\(\ds \) \(=\) \(\ds \floor {\frac {39} {19} }\)
\(\ds \) \(=\) \(\ds 2\)


Similarly:

$a_{23} = 1$
$a_{29} = 1$
$a_{31} = 1$
$a_{37} = 1$

Hence the result.

$\blacksquare$