Divisor Counting Function/Examples/60

From ProofWiki
Jump to navigation Jump to search

Example of Use of $\tau$ Function

$\tau \left({60}\right) = 12$

where $\tau$ denotes the $\tau$ Function.


Proof

From Tau Function from Prime Decomposition:

$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$

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:

$60 = 2^2 \times 3 \times 5$


Thus:

\(\ds \tau \left({60}\right)\) \(=\) \(\ds \tau \left({2^2 \times 3^1 \times 5^1}\right)\)
\(\ds \) \(=\) \(\ds \left({2 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)\)
\(\ds \) \(=\) \(\ds 12\)


The divisors of $60$ can be enumerated as:

$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$

This sequence is A018266 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

$\blacksquare$