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Let $\tau$ and $\sigma$ be the arithmetic functions defined by

  • $\tau(n)$ is the number of divisors of n, i.e $\tau(n) = \sum_{d\mid n}1$,

  • $\sigma(n)$ is the sum of divisors of n , i.e $\sigma (n) = \sum_{d\mid n}d$,

for all n.

What is $(\sigma*\tau)(12)$? (The Dirichlet product)

I know this gives $\sum_{d\mid 12}\sigma(d)\tau(12/d)$ but do not know where to go from here.

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    The divisors $d\mid 12$ are $d=1,2,3,4,6,12$. Now just evaluate each term of the sum. – 2017-02-16
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    Since $\sigma$ and $\tau$ are multiplicative, so is $\sigma*\tau$, so you only need to compute $(\sigma*\tau)(3)$ and $(\sigma*\tau)(4)$ – 2017-02-16

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