Divisor Count of 945
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Example of Use of Divisor Count Function
- $\map {\sigma_0} {945} = 16$
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:
- $945 = 3^3 \times 5 \times 7$
Thus:
\(\ds \map {\sigma_0} {945}\) | \(=\) | \(\ds \map {\sigma_0} {3^3 \times 5^1 \times 7^1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {3 + 1} \paren {1 + 1} \paren {1 + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 16\) |
The divisors of $945$ can be enumerated as:
- $1, 3, 5, 7, 9, 15, 21, 27, 35, 45, 63, 105, 135, 189, 315, 945$
This sequence is A018736 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
$\blacksquare$