Smallest Consecutive Even Numbers such that Added to Divisor Count are Equal

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Theorem

$30$ is the smallest positive even integer $n$ with the property:

\(\ds n + \map {\sigma_0} n\) \(=\) \(\ds m\)
\(\ds \paren {n + 2} + \map {\sigma_0} {n + 2}\) \(=\) \(\ds m\)
\(\ds \paren {n + 4} + \map {\sigma_0} {n + 4}\) \(=\) \(\ds m\)

where:

$m \in \Z_{>0}$ is some positive integer
$\map {\sigma_0} n$ is the divisor count function: the number of divisors of $n$.

In this case, where $n = 30$, we have that $m = 38$.


Proof

From Divisor Count Function from Prime Decomposition, we have:

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

where the prime decomposition of $n$ is:

$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$


\(\ds 2 + \map {\sigma_0} 2\) \(=\) \(\ds 2 + 2\) \(\ds = 4\) as $2 = 2^1$
\(\ds 4 + \map {\sigma_0} 4\) \(=\) \(\ds 4 + 7\) \(\ds = 7\) as $4 = 2^2$
\(\ds 6 + \map {\sigma_0} 6\) \(=\) \(\ds 6 + 4\) \(\ds = 10\) as $6 = 2^1 3^1$
\(\ds 8 + \map {\sigma_0} 8\) \(=\) \(\ds 8 + 4\) \(\ds = 12\) as $8 = 2^3$
\(\ds 10 + \map {\sigma_0} {10}\) \(=\) \(\ds 10 + 4\) \(\ds = 14\) as $10 = 2^1 5^1$
\(\ds 12 + \map {\sigma_0} {12}\) \(=\) \(\ds 12 + 6\) \(\ds = 18\) as $12 = 2^2 3^1$
\(\ds 14 + \map {\sigma_0} {14}\) \(=\) \(\ds 14 + 4\) \(\ds = 18\) as $14 = 2^1 7^1$
\(\ds 16 + \map {\sigma_0} {16}\) \(=\) \(\ds 16 + 5\) \(\ds = 21\) as $16 = 2^4$
\(\ds 18 + \map {\sigma_0} {18}\) \(=\) \(\ds 18 + 6\) \(\ds = 24\) as $18 = 2^1 3^2$
\(\ds 20 + \map {\sigma_0} {20}\) \(=\) \(\ds 20 + 6\) \(\ds = 26\) as $20 = 2^2 5^1$
\(\ds 22 + \map {\sigma_0} {22}\) \(=\) \(\ds 22 + 4\) \(\ds = 26\) as $22 = 2^1 11^1$
\(\ds 24 + \map {\sigma_0} {24}\) \(=\) \(\ds 24 + 8\) \(\ds = 32\) as $24 = 2^3 3^1$
\(\ds 26 + \map {\sigma_0} {26}\) \(=\) \(\ds 26 + 4\) \(\ds = 30\) as $26 = 2^1 13^1$
\(\ds 28 + \map {\sigma_0} {28}\) \(=\) \(\ds 28 + 6\) \(\ds = 34\) as $28 = 2^2 7^1$
\(\ds 30 + \map {\sigma_0} {30}\) \(=\) \(\ds 30 + 8\) \(\ds = 38\) as $30 = 2^1 3^1 5^1$
\(\ds 32 + \map {\sigma_0} {32}\) \(=\) \(\ds 32 + 6\) \(\ds = 38\) as $32 = 2^5$
\(\ds 34 + \map {\sigma_0} {34}\) \(=\) \(\ds 34 + 4\) \(\ds = 38\) as $34 = 2^1 17^1$

$\blacksquare$


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