Sequence of 9 Consecutive Integers each with 48 Divisors

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Theorem

The $9$ integers beginning $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 044$ each has $48$ divisors.


Proof

In the below, $\sigma_0$ denotes the divisor count function.


\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 044}\) \(=\) \(\ds 48\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 044$
\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 045}\) \(=\) \(\ds 48\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 045$
\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 046}\) \(=\) \(\ds 48\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 046$
\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 047}\) \(=\) \(\ds 48\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 047$
\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 048}\) \(=\) \(\ds 48\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 048$
\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 049}\) \(=\) \(\ds 48\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 049$
\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 050}\) \(=\) \(\ds 48\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 050$
\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 051}\) \(=\) \(\ds 48\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 051$
\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 052}\) \(=\) \(\ds 48\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 052$


but then:

\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 043}\) \(=\) \(\ds 32\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 043$
\(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 053}\) \(=\) \(\ds 8\) $\sigma_0$ of $17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 053$

$\blacksquare$


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