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Let's say I sum up all of the numbers from 1 to $n$ (we'll say $n$ is 6 for this example). In this case, I'd get a total of 21 ($1+2+3+4+5+6=21$). Now let us pose the following question: if I replaced the $+$s with $\times$s, what would I have to divide/multiply each element in the sequence by to get the same answer as with addition? As in, if

$$f_1(x)=1+2+3+...+x$$

then what would $y$ be at each point in

$$f_2(x)=\frac{1}{y_1}\times\frac{2}{y_2}\times\frac{3}{y_3}...\times\frac{x}{y_x}$$

I started calculating some of the elements of $y$, and (if I didn't make a mistake), I got:

$$\begin{array}{c|c|c|c|} n & \text{Total ($f_1$)} & \text{Fraction of $y_n$} & \text{Decimal of $y_n$} \\ \hline \text{1} & 1 & 1 & 1\\ \hline \text{2} & 3 & \frac{2}{3} & 0.66\dot6\\ \hline \text{3} & 6 & \frac{3}{2} & 1.5\\ \hline \text{4} & 10 & \frac{12}{5} & 2.4 \\ \hline \text{5} & 15 & \frac{10}{3} & 3.33\dot3 \\ \hline \text{6} & 21 & \frac{30}{7} & 4.286... \\ \hline \end{array}$$

Rather, we could get the value of $y_n$ at any point using the following formula (provided a positive integer $n$, $1<n$):

$$y_n=\frac{(\sum_{i=0}^{n-1} i)\times n}{\sum_{i=0}^{n} i}$$

The question is, is there a more effective way to calculate this - the above formula is essentially doing the same as I did by hand (adding up all of the numbers, dividing, etc.), but... as a formula. I'm also aware of Gauss's method of summing numbers, but I'm curious to find out if there's some deeper mathematical link between these numbers - possible a connection to some sequence on OEIS?

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  • $\begingroup$ So $$f_2(x)=\prod_{n=1}^{x}\frac{n}{y_n}=\frac{x!}{\prod_{n=1}^{x}y_n}\ .$$ Also note that $$\sum_{i=0}^{n}i=\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\ .$$ Are you trying to find $y_k$? What exactly are you asking? $\endgroup$
    – clathratus
    May 29, 2019 at 16:44
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    $\begingroup$ $\sum_{i=0}^n i = \frac{n(n+1)}{2}.$ So you get $$y_n=\frac{(n-1)n^2}{n(n+1)}=\frac{n(n-1)}{n+1}=n-2+\frac{2}{n+1}$$ for $n>1.$ $\endgroup$
    – Thomas Andrews
    May 29, 2019 at 16:48
  • $\begingroup$ @clathratus correct. $\endgroup$
    – Geza Kerecsenyi
    May 29, 2019 at 17:48

1 Answer 1

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We have that $\sum_{i=0}^n i = \frac{n(n+1)}{2}.$

So you get $$y_n=\frac{n\frac{n(n-1)}{2}}{\frac{n(n+1)}{2}}=\frac{n(n-1)}{n+1}=n-2+\frac{2}{n+1}$$ for $n>1.$

As for the significance of this sequence, probably not much.

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  • $\begingroup$ For some reason, when I try to plot this in Desmos it's returning undefined: i.stack.imgur.com/hFkwS.png $\endgroup$
    – Geza Kerecsenyi
    May 29, 2019 at 17:00
  • $\begingroup$ The denominator $n-2+\frac{2}{n+1}$ is zero when $n=1.$ You want $\prod_{n=2}^x,$ since the formula only works for $n>1$ and $y_1=1.$ @GezaKerecsenyi $\endgroup$
    – Thomas Andrews
    May 29, 2019 at 17:03
  • $\begingroup$ That's fixed it now, thanks. $\endgroup$
    – Geza Kerecsenyi
    May 29, 2019 at 17:04
  • $\begingroup$ Actually, @ThomasAndrews, I'll keep the question open just for a while to see if anyone can give any significance of the series (given the title of the question). If (as I suppose you are), you are correct in that there is no significance, I'll re-mark it as the answer. I've upvoted though, for now. $\endgroup$
    – Geza Kerecsenyi
    May 29, 2019 at 17:08
  • $\begingroup$ Also, I've found something pretty fun about this (though likely unrelated): if you extend the formula to an infinite series (like i.stack.imgur.com/5pMEz.png), you get some pretty interesting patterns, such as all of the odd numbers (using $x=2$), and some other fun things. $\endgroup$
    – Geza Kerecsenyi
    May 29, 2019 at 17:39

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