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If a strongly additive arithmetic function $g$ is supported on primes up to $y$, its $j$th uniform central moment differs from the independent Bernoulli random variables model with probabilities $1/p$ by $O(j{y}^j({\sum }_{p\le y}|g(p)|{)}^j/N)$. Expansion of divisibility indicator functions reduces the comparison to $\lfloor N/d\rfloor /N=1/d+O(1/N)$.

For a real strongly additive arithmetic function with uniformly bounded values on primes, put $A(x)={\sum }_{p\le x}f(p)/p$ and $B(x{)}^2={\sum }_{p\le x}f(p{)}^2/p$. If $B(x)\to \infty$, then $(f(n)-A(N))/B(N)$ under uniform sampling from $[N]$ converges to the standard normal distribution. For the prime omega function, the mean and variance scales are $\log \log N$.

One general form of the Erdős-Kac theorem is the following. Let $f$ be a real strongly additive arithmetic function, with $|f(p)|\le C$ for all primes, and defineIf $B(x)\to \infty$, then under the discrete uniform distribution on $[N]$,Here the arrow denotes convergence in distribution and $\mathcal{N}(0,1)$ is the standard normal distribution. Equivalently, the probability of being at most $z$ tends to the standard normal distribution function $\Phi (z)$ for every real $z$. Replacing $B(x{)}^2$ by ${\sum }_{p\le x}f(p{)}^2(1-1/p)/p$ is equivalent, because the difference is bounded by $C^2{\sum }_p{p}^{-2}$.

This is a general bounded-prime-weight form, not just the special case $f=\omega$. In that case the Mertens theorem for reciprocal primes gives $A(N)=\log \log N+O(1)$ and $B(N{)}^2=\log \log N+O(1)$, so the classical prime-factor central limit theorem is

Write $y=N^{1/\phi (N)}$, $L={\sum }_{p\le y}|g(p)|$, and $A={\sum }_{p\le y}g(p)/p$. Put $I_p(n)=1_{p\mid n}$ and $Y_p=I_p-1/p$. The strongly additive arithmetic function identity is $g-A={\sum }_{p\le y}g(p)Y_p$.

For any tuple $p_1,\dots ,p_j$, expand ${\prod }_{r=1}^j(I_{p_r}-1/p_r)$ by choosing an indicator in some positions and the constant in the others. If the selected positions contain distinct primes whose product is $d$, thenFor the corresponding product of the independent Bernoulli random variables $X_p$, the expectation is exactly $1/d$. This includes repeated primes, since their indicator functions are idempotent, and the empty selection has zero error. Summing absolute values of all expansion coefficients bounds the difference bywhen $y\ge 2$. Expanding the $j$th power and summing the absolute coefficient weights gives

It remains to center at the actual expectation. Since ${\mathbb{E}}_N{I}_p=\lfloor N/p\rfloor /N$, we have $|{\mathbb{E}}_{N}g-A|\le L/N$. For each $n$, both $g(n)-A$ and $g(n)-{\mathbb{E}}_{N}g$ have absolute value at most $L$, because all their indicator coefficients have absolute value at most one. The identity $u^j-v^j=(u-v){\sum }_{r=0}^{j-1}{u}^{j-1-r}{v}^r$ therefore bounds the change in the central moment by $j{L}^j/N$.

Finally, independence factors each term of the independent central moment. If $q_1,\dots ,q_w$ are the distinct primes of the tuple and $a_i$ their multiplicities, the desired comparison becomesIf $y<2$, then $g=0$ and both sides vanish, so the formula remains valid. The argument also covers complex coefficients $g(p)$; a real-valued assumption is needed for the later normal distribution conclusion, not for this centered moment comparison for additive arithmetic functions.

The prime omega function $\omega (n)$ counts the distinct prime factors of a positive integer $n$, so it is the cardinality of the prime support of an integer $P(n)$. For a prime factorization with distinct prime numbers $p_1,\dots ,p_r$ and positive exponents,At $n=1$ there are no prime factors, so the empty sum gives $\omega (1)=0$. The total number of prime factors instead gives $\Omega (n)={\sum }_i{a}_i$. ^[ref]

For $72=2^3\cdot 3^2$, the prime support of an integer is $\{2,3\}$ and $\omega (72)=2$. The comparison examples give $\omega (30)=3$, $\omega (32)=1$ and $\omega (36)=2$. Raising an existing prime factor to a higher power does not affect the prime omega function: $\omega (p^k)=1$ for a prime number $p$ and $k\ge 1$. Thus the prime omega function is a strongly additive arithmetic function, withThe reason is that coprime integers have disjoint prime supports of an integer. For arbitrary positive integers, their prime support of an integer combines by set union, and the shared prime factors are exactly those of the greatest common divisor. Counting this set union gives the more general identityFor $a=12$, $b=18$, both prime supports of an integer are $\{2,3\}$ and $\gcd (a,b)=6$, so $\omega (216)=2+2-2=2$. In particular $\omega (4)=1\ne 2=\omega (2)+\omega (2)$: the prime omega function is not a completely additive arithmetic function.

The prime omega function is useful for counting choices involving distinct prime factors. Every squarefree divisor of $n$ chooses a subset of its prime support of an integer, so there are $2^{\omega (n)}$ squarefree divisors. For $72$, this gives $2^2=4$, namely $1,2,3,6$. Also $\omega (\mathrm{rad}(n))=\omega (n)$, because the radical of an integer keeps exactly that prime support of an integer. The total number of prime factors records what this count discards, and the two counts agree exactly on squarefree integers.

A strongly additive arithmetic function satisfies $f(p^k)=f(p)$ for every prime $p$ and $k\ge 1$, so $f(n)={\sum }_{p\mid n}f(p)$. The prime omega function is an example.