Sequence of Consecutive Integers with Same Number of Divisors

Theorem

The following sequence of consecutive integers all have the same number of divisors, that is, $8$:

$40 \, 311, 40 \, 312, 40 \, 313, 40 \, 314, 40 \, 315$

This is the longest such sequence known.

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

$\ds \map {\sigma_0} {40 \, 311}$

$=$

$\ds 8$

$\ds \map {\sigma_0} {40 \, 312}$

$=$

$\ds 8$

$\ds \map {\sigma_0} {40 \, 313}$

$=$

$\ds 8$

$\ds \map {\sigma_0} {40 \, 314}$

$=$

$\ds 8$

$\ds \map {\sigma_0} {40 \, 315}$

$=$

$\ds 8$

Then we have:

$\ds \map {\sigma_0} {40 \, 310}$

$=$

$\ds 16$

$\ds \map {\sigma_0} {40 \, 316}$

$=$

$\ds 6$

$\blacksquare$