Asymptotics of $p_k$-adic valuation of the sum of the divisors of the $n$-th primorial

Question

Given this product:

$$a(n) = \prod_{k=1}^{n} (1+p_k)$$

where $p_k$ is the $k$-th prime number and which can be interpreted also as the sum of the divisors of the $n$-th primorial (OEIS A054640), is there a way to get an asymptotic estimate for $n \to \infty$ of the $p_k$-adic valuation of $a(n)$:

$$\nu_{p_k}(a(n))$$?

As expected, lower primes seem more probable:

\begin{array}{cccc} n&\nu_2(a(n))&\nu_3(a(n))&\nu_5(a(n))\\ 1&0&1&0\\ 2&2&1&0\\ 3&3&2&0\\ 4&6&2&0\\ 5&8&3&0\\ 6&9&3&0\\ 7&10&5&0\\ 8&12&5&1\\ 9&15&6&1\\ 10&16&7&2 \end{array}

After @Denis Shatrov first comment, and some numerical tests for $n \le 1000000$ here are some guesses:

$$\nu_2(a(n)) \approx 2n$$

$$\nu_3(a(n)) \approx \frac{3}{4}n$$

$$\nu_5(a(n)) \approx \frac{5}{16}n$$

$$\nu_7(a(n)) \approx \frac{7}{36}n$$

Answer 1

Using more standard notation: for any fixed prime $q$, \begin{align*} v_q\biggl( \prod_{p\le x} (p+1) \biggr) &= \sum_{p\le x} v_q(p+1) \\ &= \sum_{p\le x} \sum_{\substack{k\ge 1 \\ q^k \mid (p+1)}} 1 \\ &= \sum_{k\ge1} \sum_{\substack{p\le x \\ p\equiv-1\pmod{q^k}}} 1 \\ &= \sum_{k\ge1} \pi(x;q^k,-1). \end{align*} Since $\pi(x;m,a) \sim \frac1{\phi(m)} \frac x{\log x}$ for every $a$ with $\gcd(a,m)=1$, we deduce that $$ v_q\biggl( \prod_{p\le x} (p+1) \biggr) \sim \sum_{k\ge1} \frac1{\phi(q^k)} \frac x{\log x} = \sum_{k\ge1} \frac1{q^{k-1}(q-1)} \frac x{\log x} = \frac{q}{(q-1)^2} \frac x{\log x}. $$

Answer 2

The post is to determine an error term for Greg Martin's formula valid for all prime $q$. From his answer, we know that

$$ V(x):=\nu_q\left(\prod_{p\le x}(1+p)\right)=\sum_{k\ge1}\pi(x;q^k,-1).\tag1 $$

The convergence of (1) is of no issue as $\pi(x;q^k;-1)$ when $q^k>x$. By the prime number theorem for arithmetic progression, we have

$$ \pi(x;m,a)={1\over\phi(m)}\operatorname{li}(x)+E(x;m,a),\tag2 $$

where $\operatorname{li}(x)$ is the logarithmic integral and by Siegel-Walfisz theorem, for all $A>0$ there exists some $C_A>0$ such that

$$ |E(x;m,a)|\le{C_Ax\over(\log x)^{A+1}} $$

uniformly for all $x\ge2$, $m\ge1$, and $(a,m)=1$, so plugging (2) into (1) gives

$$ \begin{aligned} \left|V(x)-\sum_{q^k\le x}{\operatorname{li}(x)\over\phi(q^k)}\right| &\le\sum_{q^k\le x}|E(x;q^k,-1)| \\ &\le{C_Ax\over(\log x)^{A+1}}\sum_{k\le\log_qx}1\le{C_Ax\over(\log q)(\log x)^{A}}. \end{aligned}\tag3 $$

In addition, because

\begin{aligned} \sum_{q^k>x}{1\over\phi(q^k)} &={1\over q-1}\sum_{k=\lfloor\log_qx\rfloor+1}^\infty{1\over q^{k-1}}={1\over q-1}\cdot{q^{-\lfloor\log_qx\rfloor}\over1-q^{-1}} \\ &={q^{-(\lfloor\log_qx\rfloor+1)}\over(1-q^{-1})^2}<{1\over x(1-q^{-1})^2} \end{aligned}

and

$$ \sum_{k\ge1}{1\over\phi(q^k)}={1\over q-1}\sum_{k\ge1}{1\over q^{k-1}}={q\over(q-1)^2}, $$

we convert (3) into

$$ \begin{aligned} \left|V(x)-{q\over(q-1)^2}\operatorname{li}(x)\right| &<{\operatorname{li}(x)\over x(1-q^{-1})^2}+{C_Ax\over(\log q)(\log x)^A}. \\ &\le{4\operatorname{li}(x)\over x}+{C_Ax\over(\log2)(\log x)^A}. \end{aligned} \tag4 $$

Because $\operatorname{li}(x)/x\sim1/\log x\to0$, we conclude from (4) that for all $A>0$, there exists some $B_A>0$ such that for all $x\ge2$ and prime $q$, we have

$$ V(x)={q\over(q-1)^2}\operatorname{li}(x)+\vartheta{B_Ax\over(\log x)^A}.\quad(|\vartheta|<1) $$