Only finitely many values of the symmetric functions of $1/1,1/2,\ldots,1/n$ are $2$-adic integers (?) For integers $n \geq k \geq 1$ let
$$H(n,k) := \sum_{1 \leq i_1 < \cdots < i_k \leq n} \frac1{i_1 \cdots i_k}$$
be the $k$-th elementary symmetric function of $\tfrac1{1},\tfrac1{2}, \ldots, \tfrac1{n}$.
Erdős and Niven [1] proved that $H(n, k)$ is an integer only for finitely many $n$ and $k$, and subsequently Chen and Tang [2] showed that $H(1,1)$ and $H(3,2)$ are the only integral values.
My question is: "Is it true that $H(n,k)$ is a $2$-adic integer only for finitely many $n$ and $k$?"
Note the two extremal cases: $H(n, 1) = 1 + \frac1{2} + \cdots + \frac1{n}$, the $n$-th harmonic number, which is well-known to be a $2$-adic integer only for $n = 1$; and $H(n,n) = 1 / n!$, which obviously is a $2$-adic integer only for $n = 1$. Note also that the $p$-adic valuation of $H(n,k)$ has been studied in [3].
Of course one may ask the more general question: "Given a prime number $p$, is it true that $H(n,k)$ is a $p$-adic integer only for finitely many $n$ and $k$?" However, an old and still open conjecture of Eswarathasan and Levine [4] states that for any prime $p$ the harmonic number $H(n,1)$ is a $p$-adic integer only for finitely many positive integer $n$. Hence, this latter question seems to be too difficult for the current methods.
[1] P. Erdős and I. Niven, Some properties of partial sums of the harmonic series, Bull. Amer. Math. Soc., 52 (1946), 248–251.
[2] Y.-G. Chen and M. Tang, On the elementary symmetric functions of $1, 1/2, . . . , 1/n$, Amer. Math. Monthly, 119 (2012), 862–867.
[3] P. Leonetti and C. Sanna, On the p-adic valuation of Stirling numbers of the first kind, Acta Mathematica Hungarica 151 (2017), 217–231.
[4] A. Eswarathasan and E. Levine, $p$-integral harmonic sums, Discrete Math., 91 (1991), 249–257.
 A: The following is only a partial answer. The number $H(n,k)$ is not a $2$-adic integer "for most $n$". I will stick to the case $k=2$ for convenience. I claim that there is a sequence $(a_j)_{j \geq 0} \in \{ 0,1 \}^{\mathbb{N}}$, with $a_0 =1$, such that if $n = \sum_{j=0}^r b_j 2^{r-j}$ is the binary expansion of an integer $n$, with $b_0 = 1$, and if $(b_j)_{j=0}^r \neq (a_j)_{j=0}^r$, then
$$
v_2(H(n,2)) = - 2r + s,
$$
where $s$ is the smallest integer such that $b_s \neq a_s$. In particular, $H(n,2)$ is not a $2$-adic integer, unless $n$ has the form $n = n_r = \sum_{j=0}^r a_j 2^{r-j}$ for some $r$. As $n_r \in [2^r,2^{r+1}[$, these possible counterexamples are very sparse, and this what I meant by "for most $n$" above ($n_r$ is the only possible exception in $[2^r,2^{r+1}[$).
Construction of $(a_j)_j$ : one constructs by induction sequences $(a_j,n_j,x_j)_{j \geq 0}$ such that $x_j = 2^{j-1} H(n_j,2)$ is a $2$-adic integer. One sets $a_0 = n_0 =1, x_1 =0$, and the induction step is given by


*

*$a_{j+1} \in \{ 0,1 \}$ is such that $a_{j+1} \equiv a_j + x_j \pmod 2$.

*$n_{j+1} = 2n_j + a_{j+1}$

*$x_{j+1} = 2^{j} H(n_{j+1},2)$


For example $n_0=1$, $n_1 = 3$, $n_2 = 6$, $n_3=13$, $n_4=27$, $n_5=54$, $n_6=109$, $n_7=219$. Correspondingly, $(a_j)_j = 1,1,0,1,1,0,1,1,...$
I do not know if $H(n_j,2)$ is a $2$-adic integer for only finitely many $j$ (for $j \leq 7$ this is a $2$-adic integer only for $j=0,1$). For example, for $j=6$, one has $v_2(H(n_6,2)) = v_2(H(109,2)) = -3$, which is unexpectedly large.
