# 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  proved that $H(n, k)$ is an integer only for finitely many $n$ and $k$, and subsequently Chen and Tang  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 .

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  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.

 P. Erdős and I. Niven, Some properties of partial sums of the harmonic series, Bull. Amer. Math. Soc., 52 (1946), 248–251.

 Y.-G. Chen and M. Tang, On the elementary symmetric functions of $1, 1/2, . . . , 1/n$, Amer. Math. Monthly, 119 (2012), 862–867.

 P. Leonetti and C. Sanna, On the p-adic valuation of Stirling numbers of the first kind, Acta Mathematica Hungarica 151 (2017), 217–231.

 A. Eswarathasan and E. Levine, $p$-integral harmonic sums, Discrete Math., 91 (1991), 249–257.

• Note that if $1\le k\le n/2$, then there is a unique term in the summation with the maximal power of $2$ in the denominator, and hence the sum is not a $2$-adic integer in those cases. (This is a generalization of the proof for $H(n,1)$.) – Greg Martin Mar 2 '17 at 17:47
• @GregMartin For $k = 2$ and $n = 6$ both the terms $1/(2 \cdot 4)$ and $1/(4 \cdot 6)$ have minimal $2$-adic valuations. – user40023 Mar 2 '17 at 18:14
• Hmmm, yes you're right, sorry. I don't know if anything can be gotten out of modifying my comment.... – Greg Martin Mar 2 '17 at 20:00
• Experimentally this seems plausible. Here are the minima $\min_{k>0} (-v_2(H(n,k))$ for $n=1,2,3,\ldots,256$: $010221132333133 (4^{11}) 34444(5^{22})4(5^9) (6^{45})3(6^{18})(7^{91})3(7^{36})8$ where $(n^k)$ indicates a string of $k$ consecutive $n$'s. The stray 3's for $n=109$ and $n=219$ both occur at $k=2$. gp code: f(n,p)=p=prod(i=1,n,1+x/i); vector(n,j,-valuation(polcoeff(p,j),2)); vector(256,n,f(n)) – Noam D. Elkies Mar 8 '17 at 2:26
• The next 255 values are all $8$ except for $n=439$, again with the minimum at $k=2$ (this time with valuation $-5$, not $-3$). – Noam D. Elkies Mar 8 '17 at 2:30

## 1 Answer

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.

• I did'nt check all the details but it seems to me that your is Corollary 2.2. of P. Leonetti and C. Sanna, On the p-adic valuation of Stirling numbers of the first kind, Acta Mathematica Hungarica 151 (2017), 217–231. researchgate.net/publication/303488838 – user40023 Mar 4 '17 at 9:22
• Oh, right! I should have checked the references you gave. – js21 Mar 4 '17 at 9:39