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here $d(n)$ counts the number of positive divisors of $n$.

I've tried 2 things:

  • Using Bell series. But then again it just showed me that the bell series of the square of a function is not the square of the bell series of the function.

  • Expand it all and use combinatorics. If: $$n = \prod_{i=1}^r p_i^{\alpha_i} $$

then if $d \vert n$:

$$d = \prod_{i=1}^rp_i^{\beta_i}, 0\leq \beta_i \leq \alpha_i$$

so

$$\sum_{t \vert n}d(t) = \sum \prod_{i=1}^r (\beta_i + 1)$$ for all possible combitions of $(\beta_1,\beta_2,...,\beta_r)$ but I couldn't develop it very much

onlyme
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1 Answers1

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You are on the right track, but a simple way is to notice that both

$$ a(n) = \sum_{t\mid n}d^3(t), \qquad b(n)=\left(\sum_{t\mid n}d(t)\right)^2 $$ are multiplicative functions, so, in order to prove $a(n)=b(n)$, it is enough to prove: $$ a(p^k) = b(p^k) $$ that is equivalent to the well-known identity: $$ \sum_{j=0}^{k}(j+1)^3 = \left(\sum_{j=0}^{k}(j+1)\right)^2 $$ since every divisor of $p^k$ is some $p^j$ with $j\in[0,k]$, and $d(p^j)=(j+1)$.

Jack D'Aurizio
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  • I'm trying to formalize that they're multiplicative, so I took $m,n$ with $(m,n) = 1$. $\sum_{t \vert mn} d^3(t)$ if $t \vert mn$ than or $t | m$ or $t | n$ than this sum would be $\sum_{t \vert m} d^3(t) + \sum_{t \vert n} d^3(t)$ than the function is additive not multiplicative, am I wrong ? – onlyme Jul 28 '15 at 21:52
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    @onlyme: that is a general fact. If $u$ is a multiplicative function, $$ v(n) = \sum_{d\mid n}u(d) $$ is a multiplicative function, too. With the Dirichlet convolution notation, $v=(1*u)$. To prove it, just use the Chinese theorem: if $\gcd(a,b)=1$, any divisor of $ab$ can be written as a divisor of $a$ times a divisor of $b$ in a unique way. – Jack D'Aurizio Jul 28 '15 at 22:44