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since ''n'' prime factors allow a sequence of binary selection (<math>p_{i}</math> or 1) from ''n'' terms for each proper divisor formed.
Clearly, <math>1 < \sigma_0(n) < n</math> for all <math>n > 2</math>, and
The divisor function is [[multiplicative function|multiplicative]],{{Why|date=May 2021|reason=the divisor function doesn't seem obviously multiplicative. What is the proof sketch?}} but not [[Completely multiplicative function|completely multiplicative]]:
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