Smooth number

Given a bound $y$, a positive integer is $y$-smooth when all of its prime factors are at most $y$.

Fix $\eta =1/k$ with $k\ge 2$. Choose $k$ primes in $[H^{\eta -{\eta }^2},H^{\eta }]$, repetitions allowed, and an integer cofactor $m$ so that $m{p}_1\cdots p_k\le H$. The product is between $H^{1-\eta }$ and $H$, so $m\le H^{\eta }$ and all resulting integers are $H^{\eta }$-smooth. The Mertens theorem for reciprocal primes makes the reciprocal prime sum tend to $-\log (1-\eta )>0$. Since $\lfloor H/(p_1\cdots p_k)\rfloor \ge H/(2p_1\cdots p_k)$, there are ${\gg }_{\eta }H$ ordered representations. Each integer has only $O_{\eta }(1)$ prime factors in that interval, counted with multiplicity, so representation multiplicities are bounded by a constant depending on $k$. This proves the density bound. For arbitrary fixed positive $\eta$, choose a smaller reciprocal-integer exponent.

// Target: number-theory.bigb

For a specified finite prime factor base $B$, a nonzero integer is $B$-smooth if every prime dividing its absolute value belongs to $B$. Negative residues are handled by including the sign $-1$ in their exponent vectors. This notation refers to a finite prime set, rather than the alternative convention in which a numerical smoothness bound is called $B$.

For a finite set $S$ of primes, an integer is $S$-smooth when every prime dividing it belongs to $S$. Bounding the largest prime factor is equivalent to restricting the integer to be $S$-smooth for a suitable finite $S$.

If $S$ is a finite set of prime numbers, every S-smooth number has a unique expression $n=a{b}^2$, where $a$ is squarefree and its prime factors lie in $S$. There are at most $2^{|S|}$ choices for $a$ and at most $\lfloor \sqrt{x}\rfloor$ choices for $b$ when $n\le x$. Consequently $f_S(x)\le 2^{|S|}\sqrt{x}$.

The largest prime factor $P^{+}(n)$ of an integer $n>1$ is the greatest prime number dividing $n$.