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Dirichlet Divisor Problem

Let $d(n)=\nu(n)=\sigma_0(n)$ be the number of Divisors of $n$ (including $n$ itself). For a Prime $p$, $\nu(p)=2$. In general,

\sum_{k=1}^n \nu(k)=n\ln n+(2\gamma-1)n+{\mathcal O}(n^\theta),

where $\gamma$ is the Euler-Mascheroni Constant. Dirichlet originally gave $\theta \approx{1/2}$. As of 1988, this had been reduced to $\theta\approx 7/22$.

See also Divisor Function

© 1996-9 Eric W. Weisstein