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I want to know the set of integer solutions to the following sum: $$\sum\limits_{n=1}^{k}\frac{n}{a_n}=1$$ For instance, the series $a_n=\frac{1}{nk}$ satisfies this for all integer $k$. So far, this is the only solution that I have come up with for a general $k$. Here are some solutions that I have for specific values of $k$:

If $k=2$, $a_1=a_2=\frac{1}{3}$ works.

If $k=3$, $a_1=a_2=\frac{1}{4}, a_3=\frac{1}{12}$ works.

If $k=4$, $a_1=a_2=\frac{1}{5}, a_3=\frac{1}{15}, a_4=\frac{1}{20}$ works.

Are there any generalizations or ways to generate new solutions?

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    $\begingroup$ You've confused $a_n$ and $1/a_n$ (and confused me through it, too). Your solution is obviously $a_n = nk$, not $a_n = \frac{1}{nk}$. $\endgroup$ – celtschk Nov 21 '17 at 19:40
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Yet another generic solution: since $$\sum_{n=1}^{k-2} \frac{n}{2^{n+1}}=1-\frac{k}{2^{k-1}},$$ we can write the fraction $\dfrac{k}{2^{k-1}}$ in the form $$\frac{k}{2^{k-1}}=\frac{k-1}{2^{k-1}}+\frac{1}{2^{k-1}} = \color{red}{\frac{k-1}{2^{k-1}}}+\color{darkviolet}{\frac{k}{k\cdot 2^{k-1}}}.$$ Then for given $k$ ($k\ge 3$) we have solution $$(a_1,a_2,\ldots,a_{k-2},a_{k-1},a_k) = \left(4,8,\ldots, 2^{k-1}, \color{red}{2^{k-1}}, \color{darkviolet}{k\cdot 2^{k-1}}\right).$$

Examples:
$k=3$: $(4,\color{red}{4},\color{darkviolet}{12})$;
$k=4$: $(4,8,\color{red}{8},\color{darkviolet}{32})$;
$k=5$: $(4,8,16,\color{red}{16},\color{darkviolet}{80})$;
$k=6$: $(4,8,16,32,\color{red}{32},\color{darkviolet}{192})$;
$\cdots$ .


In general, the set of solutions is rather wide:

$k=3$:

(2,5,30)
(2,6,18)
(2,7,14)
(2,8,12)
(2,10,10)
(2,12,9)
(2,16,8)
(2,28,7)
(3,4,18)
(3,6,9)
(3,12,6)
(3,30,5)
(4,3,36)
(4,4,12)
(4,8,6)
(5,4,10)
(5,10,5)
(5,40,4)
(6,3,18)
(6,4,9)
(6,6,6)
(6,24,4)
(8,4,8)
(8,16,4)
(10,5,6)
(12,3,12)
(12,12,4)
(14,4,7)
(15,6,5)
(20,10,4)
(30,3,10)
(36,9,4)
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Another generic solution is to choose $a_1=a_2=\ldots=a_k=\frac{k(k+1)}{2}$ Then you have $$\sum_{n=1}^{k}\frac{n}{a_n} = \frac{2}{k(k+1)}\sum_{n=1}^k n = \frac{2}{k(k+1)}\cdot\frac{k(k+1)}{2} = 1$$

For $k=2$ this gives the solution you've already found; $a_1=a_2=3$.

For $k=3$ it gives $a_1=a_2=a_3=6$.

For $k=4$ it gives $a_1 = a_2 = a_3 = a_4 = 10$.

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Let's try the case $k=3$. Write the equation $1/a_1 + 2/a_2 + 3/a_3 = 1$ as

$$ a_3 = \frac{3 a_1 a_2}{a_1 a_2 - 2 a_1 - a_2}$$ This will certainly be an integer if the denominator is $\pm 1$ or $\pm 3$. Let's try for $+1$. So our new equation is $$ a_1 a_2 - 2 a_1 - a_2 = 1 $$ which we can solve for $a_2$: $$ a_2 = \frac{2 a_1 + 1}{a_1 - 1}$$ If $t = a_1 - 1$, this is $(2t+3)/t = 2 + 3/t$. It will be an integer if $t = -3, -1, 1$ or $3$, corresponding to $(a_1,a_2,a_3) = (-2,1,-6), (0,-1,0), (2,5,30)$ and $(4,3,36)$ respectively. Of course $(0,-1,0)$ has to be excluded because it leads to division by $0$ in the original equation.

Similarly, denominator $-3$ leads to solution $(2,1,-2)$, $-1$ to $(2,3,-18)$, and $3$ to $(-4,1,-4)$, $(2,7,14)$ and $(6,3,18)$.

The method can be extended to arbitrary $k$. First solve for $a_k$:

$$a_k = \frac{ k a_1 a_2 \ldots a_{k-1}}{P(a_1, \ldots, a_{k-1})} $$ where $P$ is a polynomial of degree $1$ in each variable. Choose one of the divisors $d$ of $k$, and consider the equation $P(a_1, \ldots, a_{k-1}) = d$, which can then be solved for $a_{k-1}$, again as a rational function of $a_1, \ldots, a_{k-2}$ with denominator of degree $1$ in each of these. Choose a value for the denominator that guarantees $a_{k-1}$ is an integer ($\pm$ a divisor of the content of the numerator). Continue until you get to something of the form $a_2 = \frac{\alpha a_1 + \beta}{a_1 -1}$, where you can take $a_1 = t + 1$ for $t$ a divisor of $\alpha + \beta$.

Taking denominators always $+1$, we get the following solutions:

$$ \eqalign{ k=2: & (2,4)\cr k=3: & (2,5,30)\cr k=4: & (2,5,31,1240)\cr k=5: & (2,5,31,1241,1923550)\cr k=6: & (2,5,31,1241,1923551, 4440055831260)\cr etc.}$$

The sequence $2,5,31,1241,1923551, \ldots$ doesn't seem to be in the OEIS. Maybe it will be soon.

EDIT: Now it is: A295391.

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Unless you want to restrict to positive integer solutions, you also have solutions such as

$${1\over-1}+{2\over2}+{3\over-3}+{4\over4}+{5\over-5}+{6\over6}+{7\over7}=1$$

and

$${1\over-1}+{2\over2}+{3\over-3}+{4\over4}+{5\over-5}+{6\over6}+{7\over-7}+{8\over4}=1$$

and the obvious generalizations for odd and even $k$.

It might be of interest to count the number of solutions (with or without a positivity restriction) for the first few values of $k$, and see if that's in the OEIS. The cases $k=1$ and $2$ are easy, but $k=3$ already seems a little tricky, at least by hand. (I've tried systematically listing the positive solutions, but I keep making mistakes. A computer program might have better luck.)

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