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The Min_25 sieve computes summatory multiplicative functions by isolating weighted prime sums and then restoring composite contributions. A useful basic form applies when $f(p)={\sum }_{r=0}^d{a}_r{p}^r$ for primes $p$, the degree $d$ is small, and values on prime powers can be computed efficiently. The formulas below explain the two stages; optimized implementations differ in scheduling and cost.^[ref]

Use quotient arguments $Q_N=\{\lfloor N/i\rfloor :1\le i\le N\}$. There are $O(\sqrt{N})$ distinct values: indices at most $\sqrt{N}$ contribute that many, and larger indices yield values below $\sqrt{N}$. The floor function identity $\lfloor \lfloor N/i\rfloor /p\rfloor =\lfloor N/(ip)\rfloor$ keeps later quotients in this table.

For each exponent $r$, initialize $G_r(v)={\sum }_{m=2}^v{m}^r$. Process primes $p$ in increasing order, and eligible $v\ge p^2$ in decreasing order, applyingThis removes composites whose smallest prime factor is $p$. The subtraction inside parentheses excludes earlier primes from the possible cofactors. Decreasing $v$ ensures each smaller argument still has its pre-$p$ value. After sieving, $H(v)={\sum }_r{a}_r{G}_r(v)$ is the desired sum of $f$ over primes up to $v$.

To restore composites, let $p_j$ be the $j$th prime and let $T(x,j)$ sum $f(n)$ over $2\le n\le x$ with every prime factor greater than $p_j$. Set $T(x,j)=0$ for $x\le p_j$. Otherwise its prime contribution is $H(x)-H(p_j)$, and addUnique decomposition by smallest prime factor and its full exponent proves this recurrence: mixed numbers use the first term, pure powers use the second. With $p_0=1$, $H(1)=0$ and $f(1)=1$, the answer is $1+T(N,0)$. No complete-multiplicativity assumption is needed.

For the Euler totient function up to six, the prime contribution is $\varphi (2)+\varphi (3)+\varphi (5)=7$. The composites add $\varphi (4)+\varphi (6)=4$, and $\varphi (1)=1$ gives total twelve. This independently chosen small example checks that prime powers and mixed products are both counted once.