Von Mangoldt divisor identity (${\sum }_{d\mid n}\Lambda (d)=\log n$)

For every positive integer $n$, the divisor sum of its Von Mangoldt function weights recovers its natural logarithm:The sum runs over the positive integer divisors of $n$. For $n=1$, its only term is $\Lambda (1)=0=\log 1$. Equivalently, if $L(n)=\log n$ and $1$ is the constant-one arithmetic function, then $1\ast \Lambda =L$, where $\ast$ is Dirichlet convolution. ^[ref]

For $12=2^2\cdot 3$, the positive integer divisors are $1,2,3,4,6,12$. Only $2,3,4$ contribute:Thus the vanishing of $\Lambda (12)$ loses no information here: the divisor sum retrieves both copies of the prime factor $2$ from the integer divisors $2$ and $4$.

Proof outline: by the fundamental theorem of arithmetic, write $n={\prod }_{i=1}^r{p}_i^{a_i}$ with distinct prime numbers $p_i$ and positive exponents $a_i$. Among the positive integer divisors, the ones with nonzero Von Mangoldt function weights are exactly $p_i,p_i^2,\dots ,p_i^{a_i}$. ConsequentlyThe natural logarithm turns the prime factorization product into a sum, and the prime powers greater than one supply exactly the required multiplicities.