Does there exist an asymptotic estimate for the following sum over primes $$ \sum_{p\leq x} \frac{\tau(p-1)}{p}\;, $$ where $\tau(n)=\sum_{d|n}1$ is the divisor function?
2026-03-25 19:03:05.1774465385
sum over primes involving divisor function (variation of the Titchmarsh divisor problem)
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@almagest, I am sorry that the note was not helpful. I wrote the note and now I included more recent theorems. However, a previous version of my note should have been helpful because of the following from the note:
Let $\tau(n)=\sum\limits_{d|n} 1$ be the divisor function. We define the following constants, where $\gamma$ is the Euler-Mascheroni constant. $$ C_1 =\frac{\zeta(2)\zeta(3)}{\zeta(6)}, $$ $$ C_2 =C_1 \left(\gamma-\sum_{p}\frac{\log p}{p^2-p+1}\right) $$
For any $A>0$, we have
$$ A(x):=\sum_{p\leq x}\tau(p-1)=C_1 x + 2C_2 \mathrm{Li} (x) + B(x) $$ where $B(x) = O\left(\frac{x}{\log^A x}\right)$.
Then by partial summation,
$$ \sum_{p\leq x} \frac{\tau(p-1)}p = \int_{2-}^{x} \frac1t dA(t) = \left[ \frac{A(t)}t\right] _{2-}^x + \int_2^x \frac{A(t)}{t^2} dt $$ $$ =O\left(\frac1{\log x} \right) +\int_2^x \left(\frac{C_1}t + 2C_2 \frac{\mathrm{Li}(t)}{t^2} \right) dt + \int_2^{\infty} \frac{B(t)}{t^2} dt $$ $$ =C_1 \log x + 2C_2 \log\log x -C_1 \log 2 - 2C_2 \log\log 2 + \int_2^{\infty} \frac{B(t)}{t^2} dt +O\left(\frac1{\log x} \right). $$