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Are there known sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ for $k > 1$ subject to the constraint $\sum_{p \mid k} \frac{1}{p+1} < 1$? (The factor of +1 in the denominator and the constraint of the sum of inverse "shifted prime divisors" is bounded by 1 are both necessary.)

A related question: Suppose $(p_i)$ is a set of $n$ consecutive primes which minimizes $1 - \sum_{i = 1}^{n} \frac{1}{p_i+1} > 0$ for a given $n > 1$. Are there known bounds for $1 - \sum_{i = 1}^{n} \frac{1}{p_i + 1}$ from below in terms of $n$, e.g., $n^{-\delta n}$ for some fixed $\delta > 0$?


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You might want to ask these questions on MO if you don't get an answer here. – Matt E Oct 23 '10 at 7:09
Thanks. I posted the same questions on MO under the same header. – user02138 Oct 25 '10 at 2:55

Actually, there is a bound for $\displaystyle \frac{1}{p}$. Perhaps, this should help. You can find this result in a standard Analytic Number theory textbook.(I have taken this from Apostol's Analytic Number Theory textbook!)

Proposition: There is a constant $A$ such that $$\sum\limits_{p \leq x} \frac{1}{p}= \log\log{x} + A + \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$$ for all $x \geq 2$, where the constant $A$ is given by $$A = 1 -\log\log{2} + \int\limits_{2}^{\infty} \frac{ R(t)}{t \log^{2}{t}} \ dt$$

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@user02138: If you had considered this, then you should mention it! – anonymous Oct 23 '10 at 6:46
up vote 1 down vote accepted

My answer on MO:

If we are allowed to consider somewhat weaker bounds, both answers depend only on the number of unitary divisors of $k$, which is $2^{\omega(k)}$. By the quoted result of O. Izhboldin and L. Kurliandchik (see Fedor's and Myerson's comments here), for any set of $n$ positive integers {$a_{1}, \dots, a_{n}$} such that $\sum_{i = 1}^{n} \frac{1}{a_{i}} <1$, we have \begin{eqnarray} \sum_{i = 1}^{n} \frac{1}{a_{i}} \leq \sum_{i = 1}^{n} \frac{1}{d_{i}} = \frac{d_{n+1} - 2}{d_{n+1} -1} < 1, \end{eqnarray} where $d_{i}$ is the $i^{\text{th}}$-Euler number, which satisfies the quadratic recurrence $d_{i} = d_{1} \cdots d_{i-1} + 1$ with $d_{1} = 2$. The first few terms of the sequence are 2, 3, 7, 43, 1807.... (A000058). It is relatively straightforward to show that the aforementioned recurrence is equivalent to $d_{i} = d_{i-1}(d_{i-1}-1) + 1$, and it is known from the recurrence that $d_{i} = \lfloor \theta^{2^{i}} + \frac{1}{2} \rfloor$, where $\theta \approx 1.2640$.... Thus, \begin{eqnarray} \sum_{p \mid k} \frac{1}{p + 1} \leq \sum_{i = 1}^{\omega(k)} \frac{1}{d_{i}} = \frac{d_{\omega(k) + 1} - 2}{d_{\omega(k)+1} - 1} = \frac{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 2}{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 1} < 1. \end{eqnarray} One can also show that $d_{i} - 1 = \lfloor \vartheta^{2^{i-1}} - \tfrac{1}{2} \rfloor$, where where $\vartheta \approx 1.5979$...., so we have \begin{eqnarray} 1 - \sum_{p \mid k} \frac{1}{p + 1} \geq 1 - \frac{d_{\omega(k)+1} - 2}{d_{\omega(k)+1} - 1} = \frac{1}{d_{\omega(k)+1} - 1} = \lfloor \vartheta^{2^{\omega(k)}} - \tfrac{1}{2} \rfloor^{-1} > 0. \end{eqnarray}

Remark E. Deutsche points out that the sequence {$d_{i} - 1$} (A007018) counts the number of ordered rooted trees with out-going degree up to 2 with all leaves at top level.

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