# Is there any mathematical significance of this sequence?

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):

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

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?

• 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? May 29, 2019 at 16:44
• $\sum_{i=0}^n i = \frac{n(n+1)}{2}.$ 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 $n>1.$ May 29, 2019 at 16:48
• @clathratus correct. May 29, 2019 at 17:48

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 $$n>1.$$
• 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 May 29, 2019 at 17:03
• 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. May 29, 2019 at 17:39