Pairs of Consecutive Integers with 6 Divisors

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Theorem

The following sequence of integers are those $n$ which fulfil the equation:

$\map {\sigma_0} n = \map {\sigma_0} {n + 1} = 6$

where $\map {\sigma_0} n$ denotes the divisor count function.

That is, they are the first of pairs of consecutive integers which each have $6$ divisors:

$44, 75, 98, 116, 147, 171, 242, 243, 244, 332, \ldots$

This sequence is A049103 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

From Divisor Count Function from Prime Decomposition:

$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$

where:

$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.


\(\ds \map {\sigma_0} {44}\) \(=\) \(\ds 6\) $\sigma_0$ of $44$
\(\ds \map {\sigma_0} {45}\) \(=\) \(\ds 6\) $\sigma_0$ of $45$


\(\ds \map {\sigma_0} {75}\) \(=\) \(\ds 6\) $\sigma_0$ of $75$
\(\ds \map {\sigma_0} {76}\) \(=\) \(\ds 6\) $\sigma_0$ of $76$


\(\ds \map {\sigma_0} {98}\) \(=\) \(\ds 6\) $\sigma_0$ of $98$
\(\ds \map {\sigma_0} {99}\) \(=\) \(\ds 6\) $\sigma_0$ of $99$


\(\ds \map {\sigma_0} {116}\) \(=\) \(\ds 6\) $\sigma_0$ of $116$
\(\ds \map {\sigma_0} {117}\) \(=\) \(\ds 6\) $\sigma_0$ of $117$


\(\ds \map {\sigma_0} {147}\) \(=\) \(\ds 6\) $\sigma_0$ of $147$
\(\ds \map {\sigma_0} {148}\) \(=\) \(\ds 6\) $\sigma_0$ of $148$


\(\ds \map {\sigma_0} {171}\) \(=\) \(\ds 6\) $\sigma_0$ of $171$
\(\ds \map {\sigma_0} {172}\) \(=\) \(\ds 6\) $\sigma_0$ of $172$


\(\ds \map {\sigma_0} {242}\) \(=\) \(\ds 6\) $\sigma_0$ of $242$
\(\ds \map {\sigma_0} {243}\) \(=\) \(\ds 6\) $\sigma_0$ of $243$
\(\ds \map {\sigma_0} {244}\) \(=\) \(\ds 6\) $\sigma_0$ of $244$
\(\ds \map {\sigma_0} {245}\) \(=\) \(\ds 6\) $\sigma_0$ of $245$


\(\ds \map {\sigma_0} {332}\) \(=\) \(\ds 6\) $\sigma_0$ of $332$
\(\ds \map {\sigma_0} {333}\) \(=\) \(\ds 6\) $\sigma_0$ of $333$

$\blacksquare$


Sources