# Fascinating number theory relationship

1. Jan 5, 2016

### willr12

Recently I noticed something odd about the triangular numbers. The basic definition is
$\displaystyle\sum_{x=1}^{n}x=T_n$
A short time after playing around with $T_n$ values I discovered something very odd-another formula for triangular numbers involving the root of the sum of cubes from 1 to n:
$\sqrt{{\displaystyle\sum_{x=1}^{n}x^3}}=T_n$
I wondered to myself whether there were other formulas for the triangular numbers involving summing consecutive powers and taking the sum to the nth root, and contemplated what values the power and root might take. In order to look at this objectively, I rewrote them this way
$\sqrt[1]{{\displaystyle\sum_{x=1}^{n}x^1}}=T_n$
And
$\sqrt[2]{{\displaystyle\sum_{x=1}^{n}x^3}}=T_n$
I noticed that the roots of the calculated sum and power of the operation within the sum went 1,1 in the first equation and 2,3 in the second, respectively (Fibonacci sequence!). I decided to test subsequent values of the Fibonacci sequence, hoping to find that 5,8 yielded triangular numbers as well. The following is the equation I used:
$\sqrt[5]{{\displaystyle\sum_{x=1}^{n}x^8}}=T_n$
Upon examination of this equation, it seems that the equation isn't entirely accurate. As n increases the accuracy of the result decreases. For example:
$\sqrt[5]{{\displaystyle\sum_{x=1}^{3}x^8}}=T_3$
5.8443... = 6

$\sqrt[5]{{\displaystyle\sum_{x=1}^{4}x^8}}=T_4$
9.3733... = 10

As you can see, with constant Fibonacci values and increasing n values, the output percent error increases and stays below the desired $T_n$ value.
Next I simply generalized the equation to include all $F_n$ values.
$\sqrt[F_{2m-1}]{{\displaystyle\sum_{x=1}^{n}x^{F_{2m}}}}=T_n$
Why, you may ask, did I have $F_{2m}$ rather than simply $F_m$ ?
Upon initial calculations, I noticed that values when one inputs odd values for m for $F_m$ in the power under the radical (and therefore even m-1 for $F_{m-1}$ for the root), the output differs from when the odd/even values are switched. This why I used 1,1 then 2,3 then 5,8 rather than 1,1 then 1,2 then 2,3 etc etc. I explored this later and will come back to this.
My next order of business was to examine what happens with a constant n value and varying values of m. The results were baffling.
First of all, I examined the following:
$\displaystyle\lim_{m\rightarrow+\infty}{\sqrt[F_{2m-1}]{{\sum_{x=1}^{n}x^{F_{2m}}}}}$
with varying values of n.
Some results:
$\displaystyle\lim_{m\rightarrow+\infty}{\sqrt[F_{2m-1}]{{\sum_{x=1}^{2}x^{F_{2m}}}}}$
You may figure that perhaps the value diverges, but contrary to what I figured, the value converges somewhat quickly, to a value you might not expect.
For
$\sqrt[F_{1}]{\displaystyle\sum_{x=1}^{2}x^{F_{2}}}=3$
This is with m=1.
To simplify, I won't put the entire equation from now on with certain m values, but instead just state the m value used. I hope you trust my calculations
m = 1 $\rightarrow$ 3
m = 2 $\rightarrow$ 3
m = 3 $\rightarrow$ 3.0338....
m = 4 $\rightarrow$ 3.0639....
m = 5 $\rightarrow$ 3.0687....
m = 6 $\rightarrow$ 3.0694....
m = 7 $\rightarrow$ 3.0695....
Clearly the value converges to an asymptote about 3.069....something. Odd value.
For n = 3, the result is similar.
m = 1 $\rightarrow$ 6
m = 2 $\rightarrow$ 6
m = 3 $\rightarrow$ 5.844....
m = 4 $\rightarrow$ 5.898...
m = 5 $\rightarrow$ 5.913....
m = 6 $\rightarrow$ 5.915....
The only difference I notice between n= 2 and n = 3 is that with 3, the value starts at the desired $T_n$ and then dips below that value before coming back up to the eventual asymptote. With n = 2, it starts at $T_2$ and immediately goes up and starts toward its limit.
It turns out that 2 is he only n value that behaves this way, as all values of n above 2 behave in the way 3 does. Here's a list of calculated asymptotal limits for the first few values of n:
n = 2 $\rightarrow$ 3.06956... ($T_n=3$)
n = 3 $\rightarrow$ 5.9156... ($T_n=6$)
n = 4 $\rightarrow$ 9.4211... ($T_n=10$)
n = 5 $\rightarrow$ 13.5199.... ($T_n=15$)
As you can see the equation becomes less accurate as n increases.
Going back to what happens when you use odd m for the power and even m for the root, another interesting pattern emerges. The limit of that, which you may define as
$\displaystyle\lim_{m\rightarrow+\infty}{\sqrt[F_{2m}]{{\sum_{x=1}^{n}x^{F_{2m+1}}}}}$
Also approaches the values calculated above. But rather than approaching the limit in the same manner, they approach the asymptote from above. Example for n = 2:
m = 1 $\rightarrow$ 5
m = 2 $\rightarrow$ 3.2075...
m = 3 $\rightarrow$ 3.0845...
m = 4 $\rightarrow$ 3.0717...
Clearly they converge to the same value from opposite directions. The fully generalized expression for this relationship would then be
$\displaystyle\lim_{m\rightarrow+\infty}{\sqrt[F_{m-1}]{{\sum_{x=1}^{n}x^{F_{m}}}}}$
Connecting the two like this shows a surprising result. The values jump back and forth over the asymptotal limit, getting closer with increasing m. Where have you seen this before? Why, it's exactly the same manner with which the ratio of successive Fibonacci numbers asymptote to the golden ratio! In other words, this behaves the same way about its limit as
$\displaystyle\lim_{n\rightarrow\infty}{\frac{F_n}{F_{n-1}}}$
does.
I will attach a picture of a graph I made which depicts the asymptotal behavior of n = 2.

My next struggle is to find the significance of these asymptotal limits, which I figure to somehow be related to the golden ratio in some way because of how the behavior around the limits is similar to that of the golden ratio. I have hit a wall now because unfortunately I'm a junior in high school and have very little resources with which to work. I'm fact, I'm currently writing this in my English class while I'm supposed to be writing a paper. Any ideas as to the significance of these limits are very well appreciated as well as opinions on the equation as a whole.

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2. Jan 5, 2016

### Khashishi

Now that is really cool. From wikipedia article on Indefinite sum
https://en.wikipedia.org/wiki/Indefinite_sum#List_of_indefinite_sums
it looks like the general formula for
$\sum\limits_x {x^a} = \frac{B_{a+1}(x)}{a+1} + C$, for a >=0
where B is a Bernoulli polynomial. https://en.wikipedia.org/wiki/Bernoulli_polynomials
Indeed, $B_4(x) = B_2(x)^2 + constant$
as you have shown.

Your limit should then go to
$\lim_{n\rightarrow \infty} \sqrt[n]{\frac{B_{2n}(x)}{2n}}$
Perhaps this can be simplified.

3. Jan 5, 2016

### micromass

$$(1^{F_{2m}} + 2^{F_{2m}} + 3^{F_{2m}})^{1/F_{2m -1}} = 3^{F_{2m}/F_{2m-1}} ( (1/3)^{F_{2m}} + (2/3)^{F_{2m}} + 1)^{1/F_{2m-1}}\sim 3^{F_{2m}/F_{2m-1}}$$
Taking limits yields $3^\varphi$.

More generally:
$$\lim_{m\rightarrow +\infty} \left(\sum_{k=1}^n k^{F_{2m}} \right)^{F_{2m-1}} = n^\varphi$$

I leave it up to you to make my argument more rigorous. Very small hint on that: https://en.wikipedia.org/wiki/Generalized_mean

Last edited: Jan 5, 2016
4. Jan 5, 2016

### willr12

Outstanding!!
$2^\varphi = 3.0695...$
And etc.
Thanks!!!! More work to come