Today I stumbled upon one of Ramanujan's infinite roots with all the integers and that got me curious so I started to try to create one of my own. I started with $2=2$. Then $$2=\sqrt4$$ Then $$2=\sqrt{1+3}$$ Then $$2=\sqrt{1+\sqrt{9}}$$ Then $$2=\sqrt{1+\sqrt{2+7}}$$ Then $$2=\sqrt{1+\sqrt{2+\sqrt{3+46}}}$$ Then $$2=\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+2112}}}}$$ Then the final step I did was $$2=\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt{5+4460539}}}}}$$ Each step did equal 2 and while doing this I got very curious at to the numbers that were following each of the counting numbers in each step. By this I mean $$3$$ in $$1+3$$ and the $$7$$ in $$2+7$$ as well as the $$46,2112,$$and$$4460539$$ After trying for a long time, I failed to create a function that would give out these values when putting in $$1,2,3,4...$$ Which is why I have come here. This is purely for curiosity, but I need help to create a function that will spit out $$3,7,46,2112,4460539,..etc$$. Thanks!
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Do you want an explicit function or a recursive one? – AspiringMat Oct 28 '16 at 01:13
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Just by inverting you get $(\dotsc(((2^2-1)^2-2)^2-3)^2\dotsc)^2-n$. With some persistence and some binomial theorems you could make it more explicit. – ziggurism Oct 28 '16 at 01:15
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I'm sorry but I'm just a high school student so I don't know exactly what you mean by explicit or recursive – Brothersquid Oct 28 '16 at 01:17
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I believe the last term you've calculated is incorrect and should be $4460539$. – Noah Schweber Oct 28 '16 at 01:26
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yes I see my mistake – Brothersquid Oct 28 '16 at 01:29
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This is sequence $A198959$ at OEIS – Claude Leibovici Oct 28 '16 at 07:00
3 Answers
It is easy to see that the sequence satisfies the recurrent relation $$a_1 = 2,\quad a_{k+1} = a_k^2 - k$$ This sequence is discussed in https://mathoverflow.net/questions/147217/asymptotic-behavior-of-the-sequence-u-n-u-n-12-n, where it is given also the explicit (up to a constant) expression for it: $$a_n = \left\lceil \lambda^{2^{n-1}}\right\rceil$$ where $\lambda$ is some constant.
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I have looked at the provided link and I am having trouble making sense of it and finding the explicit formula for my problem – Brothersquid Oct 28 '16 at 01:39
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They claim that $a_n = \lceil \lambda^{2^{n+1}}\rceil$, where $\lambda$ is a certain constant. I didn't check though myself. – trosos Oct 28 '16 at 01:47
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The problem is, the only way to find $\lambda$ seems to be evaluating the terms of the original recursion. As impressive as the formula is, it's not good for this application. – Kaynex Oct 28 '16 at 01:53
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Ok, I overlooked that subtlety in the definition of $\lambda$. It however is a step towards an explicit relation. – trosos Oct 28 '16 at 01:59
$7=3^2-2$
$46=7^2-3$
$2112=46^2-4$
Do you get the pattern?
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yes I did see that pattern but I did not see a way to create a function out of that as all it says is $$f(x)=(f(x-1))^2-x$$ whatever f(x) may be – Brothersquid Oct 28 '16 at 01:23
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1If you let $f(x)$ be a function where its domain (i.e. values you can input to the function) are only integers, Then we can define it this way: $f(1)=3$ and $f(x)=f(x-1)^2-x$. This is what we call a recursive function since it doesn't give us the output explicitly in terms of $x$ but rather in terms of a smaller version of the function. So for example we have:
$f(2)=f(1)^2-2=3^2-2$, $f(3)=f(2)^2-3=...$, you get the idea.
– AspiringMat Oct 28 '16 at 01:28 -
You might wanna read on recursive functions, they are extremely useful in solving a range or problems. – AspiringMat Oct 28 '16 at 01:29
Based on what you were doing, there's a clear recursion: $$f(n) = f(n - 1)² - n$$ This is a nonhomogenous quadratic map and likely has no closed form solution. I would say you should be ready for the possibility that no "plug and play" function exists. I can work on any other type of solution for this, we'll see if anything comes up.
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