Divisor Count of 17,796,126,877,482,329,126,048

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Example of Use of Divisor Count Function

$\map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 048} = 48$

where $\sigma_0$ denotes the divisor count function.


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 $n$.


We have that:

$17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 048 = 2^5 \times 19 \times 4 \, 590 \, 338 \, 339 \times 6 \, 376 \, 424 \, 429$


Thus:

\(\ds \) \(\) \(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 048}\)
\(\ds \) \(=\) \(\ds \map {\sigma_0} {2^5 \times 19^1 \times 4 \, 590 \, 338 \, 339^1 \times 6 \, 376 \, 424 \, 429^1}\)
\(\ds \) \(=\) \(\ds \paren {5 + 1} \times \paren {1 + 1} \times \paren {1 + 1} \times \paren {1 + 1}\)
\(\ds \) \(=\) \(\ds 6 \times 2^3\)
\(\ds \) \(=\) \(\ds 48\)

$\blacksquare$