For a given positive integer $a$, let $b(a)$ be the smallest integer $b > a$ such that the denominator of $\frac{1}{a} + \frac{1}{a+1} + \ldots + \frac{1}{b}$ is larger than the denominator of $\frac{1}{a} + \frac{1}{a+1} + \ldots + \frac{1}{b+1}$. For example, $b(3) = 5$, as $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{47}{60}$ while $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{19}{20}$, and $60 > 20$.
In this paper (where the definition of $b(a)$ is shifted by $1$, but I'll ignore that for now) it is proven that $b(a)$ exists for all $a$, and we moreover have the upper bound $b(a) < 4.374a$ for all $a > 1$.
For this post I am more interested in the lower bound, however. In that regard, Section $3.3$ of the referenced paper proves $b(a) > a + (c_0 + o(1))\log(a)$ for some specific constant $c_0$ with $0.54 < c_0 < 0.55$. As it is conjectured that this lower bound is optimal (that is, for every $\epsilon > 0$ I expect there to be infinitely many $a$ for which $b(a) < a + (c_0 + \epsilon)\log(a)$ ), I think it would be worthwhile to add the decimal expansion of $c_0$ to the OEIS. As of right now I only know that it starts with $0.54$, and my best guess is that the next digit is a $6$, although it could be a $5$.
To explain how this constant $c_0$ can be calculated takes a bit of time, so please bear with me. However, I can tell you the tldr already: we need to count the number of derangements in certain large Galois groups. So if you happen to have some knowledge in that area, you are more than welcome to help!
Now, for a positive integer $d$, define the polynomial $f_d(x) := \sum_{i=0}^d \prod_{0 \le j \le d, j\neq i} (x-j)$ and let $\delta(f_d)$ be the density of primes $p$ for which $f_d(x) \equiv 0 \pmod{p}$ is solvable. We further define $c := \sum_{d=1}^{\infty} \frac{\delta(f_d)}{d(d+1)}.$ By the aforementioned Section $3.3$ we then have the following bounds on $b(a)$: $\frac{1}{1+c} \le \liminf_{a \rightarrow \infty} \frac{b(a) - a}{\log(a)} \le \frac{1}{2c}.$ The constant $c_0$ I referenced earlier is the lower bound $\frac{1}{1+c}$ here.
All in all we see that to determine $c_0$, it suffices to know $\delta(f_d)$ for all $d$.
As it turns out, $\delta(f_d) = 1$ for all odd $d$. The reason for this is that we have the equality $f_d(x) = (-1)^d f_d(d-x)$, which gives $f_d(d/2) = 0$ implying that $f_d(x) \equiv 0 \pmod{p}$ is solvable for all $p > 2$. When $d$ is even, things are less clear, and here we get to the heart of the matter.
Let $G_d$ be the Galois group of $f_d(x)$, viewed as a group of permutations on the roots of $f_d(x)$. We recall that a permutation $\sigma$ on a set $X$ is called a derangement if $\sigma(x) \neq x$ for all $x \in X$. A theorem by Frobenius then states that, if $f_d(x)$ is irreducible (which I have checked for all even $d \le 500$ using PARI/GP), then $\delta(f_d)$ is equal to the proportion of elements $\sigma \in G_d$ such that $\sigma$ is not a derangement.
So what do these Galois groups $G_d$ look like when $d = 2l$ is even? I can prove that they are subgroups of the so-called signed symmetric group $S^{+}_l$, which has order $2^l \cdot l!$. Moreover, if $G_d$ is equal to the full group $S^{+}_l$ (which I found happens for all even $d \le 60$ except $d = 8, 24, 48$), then there is a formula for the number of derangements in $G_d$ which we can then simply plug in, in order to find $\delta(f_d)$. The interesting case is when $G_d$ is a proper subgroup of $S^{+}_l$, and then I am no longer sure what's true. So the step by step plan (which may have shortcuts I don't know about) is:
- Determine $G_d$ for as many even $d$ as possible, and check that $f_d(x)$ is indeed irreducible for these $d$.
- Calculate the number of derangements in $G_d$.
- Use this (and other bounds on $\delta(f_d)$ like $\frac{1}{d} \le \delta(f_d) \le 1 - \frac{1}{d}$ that may hold in general) to find as many digits of $c_0$ as possible.
- ???
- Profit.
All steps can plausibly be carried out using packages like GAP/Sage/Magma, perhaps combined with some group-theoretic arguments.
Thanks in advance for anyone interested to help out, and thanks already for finishing this wall of text.
For a given positive integer$a$ , let $b(a)$ be the smallest integer $b > a$ such that the denominator of $\frac{1}{a} + \frac{1}{a+1} + \ldots + \frac{1}{b}$ is larger than the denominator of $\frac{1}{a} + \frac{1}{a+1} + \ldots + \frac{1}{b+1}$ . For example, $b(3) = 5$ , as $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{47}{60}$ while $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{19}{20}$ , and $60 > 20$ .
In this paper (where the definition of$b(a)$ is shifted by $1$ , but I'll ignore that for now) it is proven that $b(a)$ exists for all $a$ , and we moreover have the upper bound $b(a) < 4.374a$ for all $a > 1$ .
For this post I am more interested in the lower bound, however. In that regard, Section$3.3$ of the referenced paper proves $b(a) > a + (c_0 + o(1))\log(a)$ for some specific constant $c_0$ with $0.54 < c_0 < 0.55$ . As it is conjectured that this lower bound is optimal (that is, for every $\epsilon > 0$ I expect there to be infinitely many $a$ for which $b(a) < a + (c_0 + \epsilon)\log(a)$ ), I think it would be worthwhile to add the decimal expansion of $c_0$ to the OEIS. As of right now I only know that it starts with $0.54$ , and my best guess is that the next digit is a $6$ , although it could be a $5$ .
To explain how this constant$c_0$ can be calculated takes a bit of time, so please bear with me. However, I can tell you the tldr already: we need to count the number of derangements in certain large Galois groups. So if you happen to have some knowledge in that area, you are more than welcome to help!
Now, for a positive integer$d$ , define the polynomial $f_d(x) := \sum_{i=0}^d \prod_{0 \le j \le d, j\neq i} (x-j)$ and let $\delta(f_d)$ be the density of primes $p$ for which $f_d(x) \equiv 0 \pmod{p}$ is solvable. We further define $c := \sum_{d=1}^{\infty} \frac{\delta(f_d)}{d(d+1)}.$ By the aforementioned Section $3.3$ we then have the following bounds on $b(a)$ : $\frac{1}{1+c} \le \liminf_{a \rightarrow \infty} \frac{b(a) - a}{\log(a)} \le \frac{1}{2c}.$ The constant $c_0$ I referenced earlier is the lower bound $\frac{1}{1+c}$ here.
All in all we see that to determine$c_0$ , it suffices to know $\delta(f_d)$ for all $d$ .
As it turns out,$\delta(f_d) = 1$ for all odd $d$ . The reason for this is that we have the equality $f_d(x) = (-1)^d f_d(d-x)$ , which gives $f_d(d/2) = 0$ implying that $f_d(x) \equiv 0 \pmod{p}$ is solvable for all $p > 2$ . When $d$ is even, things are less clear, and here we get to the heart of the matter.
Let$G_d$ be the Galois group of $f_d(x)$ , viewed as a group of permutations on the roots of $f_d(x)$ . We recall that a permutation $\sigma$ on a set $X$ is called a derangement if $\sigma(x) \neq x$ for all $x \in X$ . A theorem by Frobenius then states that, if $f_d(x)$ is irreducible (which I have checked for all even $d \le 500$ using PARI/GP), then $\delta(f_d)$ is equal to the proportion of elements $\sigma \in G_d$ such that $\sigma$ is not a derangement.
So what do these Galois groups$G_d$ look like when $d = 2l$ is even? I can prove that they are subgroups of the so-called signed symmetric group $S^{+}_l$ , which has order $2^l \cdot l!$ . Moreover, if $G_d$ is equal to the full group $S^{+}_l$ (which I found happens for all even $d \le 60$ except $d = 8, 24, 48$ ), then there is a formula for the number of derangements in $G_d$ which we can then simply plug in, in order to find $\delta(f_d)$ . The interesting case is when $G_d$ is a proper subgroup of $S^{+}_l$ , and then I am no longer sure what's true. So the step by step plan (which may have shortcuts I don't know about) is:
All steps can plausibly be carried out using packages like GAP/Sage/Magma, perhaps combined with some group-theoretic arguments.
Thanks in advance for anyone interested to help out, and thanks already for finishing this wall of text.