Arithmetic Functions and Dirichlet Convolution — Examples

By Dirichlet convolution of arithmetic functions, $(\mathbf{1}\ast \mathbf{1})(n)={\sum }_{d\mid n}1$, so each entry in the third column is exactly the number of listed positive divisors of $n$. The rows $n=1,\dots ,6$ therefore give the values $1,2,2,3,2,4$.

The same divisor-counting rule applied to the listed divisors for $n=7,\dots ,12$ gives the remaining values $2,4,3,4,2,6$. By The divisor-counting function $\tau$, these are also the corresponding $\tau (n)$ values.

Thus every row of the table displays $(\mathbf{1}\ast \mathbf{1})(n)=\tau (n)$ on the range $1\le n\le 12$.

Since $\mathbf{1}(1)=1$, the inverse criterion gives $g(1)=1$. For $n>1$ the recursion is $g(n)=-{\sum }_{d\mid n,d<n}g(d)$. Therefore $g(2)=-g(1)=-1$, $g(3)=-g(1)=-1$, $g(4)=-(g(1)+g(2))=0$, $g(5)=-g(1)=-1$, and $g(6)=-(g(1)+g(2)+g(3))=1$.

Using the displayed values, $(\mathbf{1}\ast g)(1)=1$, $(\mathbf{1}\ast g)(2)=1+(-1)=0$, $(\mathbf{1}\ast g)(3)=1+(-1)=0$, $(\mathbf{1}\ast g)(4)=1+(-1)+0=0$, $(\mathbf{1}\ast g)(5)=1+(-1)=0$, and $(\mathbf{1}\ast g)(6)=1+(-1)+(-1)+1=0$. So $(\mathbf{1}\ast g)(n)=\epsilon (n)$ for $1\le n\le 6$, where $\epsilon$ is the identity of The Dirichlet-convolution identity and the constant-one function.

Dirichlet convolution is commutative, so the same finite check also gives $(g\ast \mathbf{1})(n)=\epsilon (n)$ on this displayed range.

By For $n\ge 1$ and any injective list $p:r\to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n={\prod }_{i<r}{p}_i^{v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n)=0$ for every prime $q$ outside the list, $12=2^2\cdot 3$ and $18=2\cdot 3^2$. Applying the prime-power formulas of The divisor functions arise by Dirichlet convolution gives the displayed products for $\tau (12)$, $\sigma (12)$, $\tau (18)$, and ${\sigma }_2(18)$.

Evaluating those products yields $\tau (12)=6$, $\sigma (12)=7\cdot 4=28$, $\tau (18)=6$, and ${\sigma }_2(18)=5\cdot 91=455$.

The direct divisor lists $\{1,2,3,4,6,12\}$ and $\{1,2,3,6,9,18\}$ have six elements each, and $1+2+3+4+6+12=28$ while $1+2^2+3^2+6^2+9^2+{18}^2=455$, so the prime-power formulas agree with direct computation in these cases.

By For every positive integer $n$, ${\sum }_{d\mid n,d>0}\phi (d)=n$, one has ${\sum }_{d\mid n}\phi (d)=n={\mathrm{id}}_1(n)$, where ${\mathrm{id}}_1$ is the power function of The power functions ${\mathrm{id}}_k$ and the divisor-power-sum functions ${\sigma }_k$.

Applying Classical Möbius inversion over positive divisors to the functions $f=\phi$ and $g={\mathrm{id}}_1$ gives $\phi (n)={\sum }_{d\mid n}\mu (d){\mathrm{id}}_1(n\mathrm{/}d)={\sum }_{d\mid n}\mu (d)(n\mathrm{/}d)$.

The final sum is exactly the Dirichlet convolution formula for $\mu \ast {\mathrm{id}}_1$, so $\phi =\mu \ast {\mathrm{id}}_1$.

The divisors of $16$ are $1,2,4,8,16$. By The von Mangoldt function, only the prime powers $2,4,8,16$ contribute, each with value $\log 2$, so ${\sum }_{d\mid 16}\mathrm{\Lambda }(d)=4\log 2=\log 16$.

The divisors of $30$ are $1,2,3,5,6,10,15,30$. Again by The von Mangoldt function, only $2,3,5$ contribute, so ${\sum }_{d\mid 30}\mathrm{\Lambda }(d)=\log 2+\log 3+\log 5=\log 30$.

These are exactly the values predicted by The divisor sum of von Mangoldt is the arithmetic-function logarithm on one prime power and one squarefree composite with three prime factors.

By Liouville's function, $\lambda (p^j)=(-1{)}^j$ for every $j\ge 0$. Therefore $(\mathbf{1}\ast \lambda )(p^4)={\sum }_{j=0}^4\lambda (p^j)={\sum }_{j=0}^4(-1{)}^j=1$.

The same computation gives $(\mathbf{1}\ast \lambda )(p^3)={\sum }_{j=0}^3\lambda (p^j)={\sum }_{j=0}^3(-1{)}^j=0$.

These explicit values match the square-indicator statement of The convolution $\mathbf{1}\ast \lambda$ detects perfect squares for an even and an odd exponent.

By The divisor functions arise by Dirichlet convolution, the function $\tau$ is multiplicative in the sense of Multiplicative arithmetic functions.

The same proposition gives $\tau (2)=2$ and $\tau (4)=3$. Hence $\tau (4)=3\mathrm{\ne }4=\tau (2{)}^2$, so $\tau (mn)=\tau (m)\tau (n)$ fails at the non-coprime pair $m=n=2$.

Therefore $\tau$ is a multiplicative arithmetic function that is not completely multiplicative, contradicting the refuted statement.

The cited paper of Odlyzko and te Riele proves that there exists a real number $x>1$ for which $\mathrm{\mid }M(x)\mathrm{\mid }\ge \sqrt{x}$. So the universal inequality in the Statement refuted fails at that value of $x$.

Hence the Mertens conjecture is false. This example is historical and non-load-bearing for the finite convolution arguments on the A page.