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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.

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  • $\begingroup$ 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 $H(n,1)$.) $\endgroup$ – Greg Martin Mar 2 '17 at 17:47
  • $\begingroup$ @GregMartin For $k = 2$ and $n = 6$ both the terms $1/(2 \cdot 4)$ and $1/(4 \cdot 6)$ have minimal $2$-adic valuations. $\endgroup$ – user40023 Mar 2 '17 at 18:14
  • $\begingroup$ Hmmm, yes you're right, sorry. I don't know if anything can be gotten out of modifying my comment.... $\endgroup$ – Greg Martin Mar 2 '17 at 20:00
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    $\begingroup$ 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)) $\endgroup$ – Noam D. Elkies Mar 8 '17 at 2:26
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    $\begingroup$ The next 255 values are all $8$ except for $n=439$, again with the minimum at $k=2$ (this time with valuation $-5$, not $-3$). $\endgroup$ – Noam D. Elkies Mar 8 '17 at 2:30
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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.

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  • $\begingroup$ 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 $\endgroup$ – user40023 Mar 4 '17 at 9:22
  • $\begingroup$ Oh, right! I should have checked the references you gave. $\endgroup$ – js21 Mar 4 '17 at 9:39

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