= Euler product for a smoothed divisor-square correlation
{c}
{title2=$E(z,z')=\zeta(1+z+z')H(z,z')/(\zeta(1+z)\zeta(1+z'))$}
For $\operatorname{Re}z,\operatorname{Re}z'>0$, the series $\sum_{d,e}\mu(d)\mu(e)d^{-z}e^{-z'}/[d,e]$ has local factor $1-p^{-1-z}-p^{-1-z'}+p^{-1-z-z'}$. Dividing by the displayed <zeta function> ratio leaves a <holomorphic function> $H$ near zero with $H(0,0)=1$. The local factor of $H$ is $(1-a-b+c)(1-c)/((1-a)(1-b))$, where $a=p^{-1-z}$, $b=p^{-1-z'}$, $c=p^{-1-z-z'}$, and equals one exactly at zero. With $z=(1+it)/\log X$, the <zeta function> poles produce the kernel $(1+it)(1+it')/((2+i(t+t'))\log X)$ that determines the <smooth divisor-square sieve asymptotic>.
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