Divisor Count of 496

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Example of Use of Divisor Counting Function

$\map {\sigma_0} {496} = 10$

where $\sigma_0$ denotes the divisor counting function.


Proof

From Divisor Counting 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:

$496 = 2^4 \times 31$

Thus:

\(\ds \map {\sigma_0} {496}\) \(=\) \(\ds \map {\sigma_0} {2^4 \times 31^1}\)
\(\ds \) \(=\) \(\ds \paren {4 + 1} \paren {1 + 1}\)
\(\ds \) \(=\) \(\ds 10\)


The divisors of $496$ can be enumerated as:

$1, 2, 4, 8, 16, 31, 62, 124, 248, 496$

This sequence is A018487 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

$\blacksquare$