Divisor Counting Function/Examples/105

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Example of Use of $\tau$ Function

$\map \tau {105} = 8$

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


Proof

From Tau Function from Prime Decomposition:

$\ds \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$

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:

$105 = 3 \times 5 \times 7$


Thus:

\(\ds \map \tau {105}\) \(=\) \(\ds \map \tau {3^1 \times 5^1 \times 7^1}\)
\(\ds \) \(=\) \(\ds \paren {1 + 1} \paren {1 + 1} \paren {1 + 1}\)
\(\ds \) \(=\) \(\ds 8\)


The divisors of $105$ can be enumerated as:

$1, 3, 5, 7, 15, 21, 35, 105$

This sequence is A018286 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

$\blacksquare$