Simple formula for $H_n = m + \alpha $?

Question

Let $H_i$ be the $i$ th harmonic number. For a given positive integer $m$ we want to find the smallest possible positive integer value $n$ such that $H_n = m + \alpha $, where $\alpha > 0$.

Let us define the solution as $f(m) = n$

Let $\operatorname{floor}$ be defined as rounding downwards. Examples $\operatorname{floor}(3,14) = 3 , \operatorname{floor}(9,99999) = 9 $)

Let $\gamma$ be the Euler-Mascheroni constant.

Now it appears that $f(m) $ has a simple closed form :

$$ f(m) = \operatorname{floor}\left( \exp(n - \gamma) - \frac{1}{2} \right) $$

How to prove this ? I assume it is necessary to assume $\gamma$ is irrational ?

I toyed around with irrationality measure and asymptotics to the digamma function. But nothing worked.

Maybe use Fourier series ?

Answer 1

I wrote a program which calculated a few first values $f(m)$ and plug them to OEIS. The search query point me to sequence A002387, and your question turned out to be well-known. I copied below the comments from OEIS

For $k\ge 1$, $$\log(k + 1/2) + \gamma < H_k< \log(k + 1/2) + \gamma + 1/(24k^2),$$ where $\gamma$ is Euler's constant (A001620). It is likely that the upper and lower bounds have the same floor for all $k\ge2$, in which case $a(n) = \lfloor \exp(n-\gamma) + 1/2\rfloor$ for all $n\ge 0$. This remark is based on a simple heuristic argument. The lower and upper bounds differ by $1/(24k^2)$, so the probability that there's an integer between the two bounds is $1/(24k^2)$. Summing that over all $k\ge 2$ gives the expected number of values of $k$ for which there's an integer between the bounds. That sum equals $\pi^2/144 - 1/24 \approx 0.02687$. That's much less than $1$, so it is unlikely that there are any such values of $k$. – Dean Hickerson, Apr 19 2003

So, if the conjecture fails then a counterexample is expected to be at the beginning of the infinity. :-) Robert G. Wilson v, and T. D. Noe provided a table of $n$, $a(n)$ for $n = 0..2303$, so if the conjecture holds for these $n$ then the heuristic probability of its failure falls to about $1/(24\cdot 2303)$.

Referring to A118050 and A118051, using a few terms of the asymptotic series for the inverse of $H(x)$, we can get an expression which, with greater likelihood than mentioned above, should give $a(n)$ for all $n\ge 0$. For example, using the same type of heuristic argument given by Dean Hickerson, it can be shown that, with probability $> 99.995%$, we should have, for all $n\ge 0$, $$a(n) = \lfloor u + 1/2 - 1/(24u) + 3/(640u^3)\rfloor$$ where $u = e^{n - \gamma}$. - David W. Cantrell (DWCantrell(AT)sigmaxi.net)