Divisor Count of 490

From ProofWiki
Jump to navigation Jump to search

Example of Use of Divisor Count Function

$\map {\sigma_0} {490} = 12$

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:

$490 = 2 \times 5 \times 7^2$

Thus:

\(\ds \map {\sigma_0} {490}\) \(=\) \(\ds \map {\sigma_0} {2^1 \times 5^1 \times 7^2}\)
\(\ds \) \(=\) \(\ds \paren {1 + 1} \paren {1 + 1} \paren {2 + 1}\)
\(\ds \) \(=\) \(\ds 12\)


The divisors of $490$ can be enumerated as:

$1, 2, 5, 7, 10, 14, 35, 49, 70, 98, 245, 490$

This sequence is A018483 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

$\blacksquare$