Divisor Count of 248

From ProofWiki
Jump to navigation Jump to search

Example of Use of Divisor Count Function

$\map {\sigma_0} {248} = 8$

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:

$248 = 2^3 \times 31$

Thus:

\(\ds \map {\sigma_0} {248}\) \(=\) \(\ds \map {\sigma_0} {2^3 \times 31^1}\)
\(\ds \) \(=\) \(\ds \paren {3 + 1} \paren {1 + 1}\)
\(\ds \) \(=\) \(\ds 8\)


The divisors of $248$ can be enumerated as:

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

This sequence is A018355 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

$\blacksquare$