5

Can I have a proof that this number exists?

The number:

$$\binom{1}{\binom{2}{\binom{3}{\binom{4}{\vdots}}}}$$

If the number exists, then what is the closed form of that number?

  • Could you be more precise here at all? I'm not sure what that means. – Marcus M Jul 17 '15 at 01:13
  • Even if you mean $\binom{n}{\binom{n-1}{\binom{n-2}{\vdots}}}$, then $\binom{k}{k-1}=k$ so it would just be $n$. – Alex R. Jul 17 '15 at 01:18
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    @bburGsamohT but couldn't you also apply that logic to say $\binom{2}{\binom{3}{\vdots}} = 0$? That would yield $\binom{1}{0} = 1$ as the answer. – Marcus M Jul 17 '15 at 01:21
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    @MarcusM my apologies, I thought it read along the lines of $\binom{\binom{a}{b}}{c}$ instead of $\binom{a}{\binom{b}{c}}$, which gave me a different interpretation! – TomGrubb Jul 17 '15 at 01:25
  • It is already in closed form, enclosed by a so many parentheses. :D – P Vanchinathan Jul 17 '15 at 02:26

3 Answers3

10

If we interpret this as $\displaystyle\lim_{n\to \infty}\,a_n$, where $a_n:=\left(\substack{{1}\\{\phantom{a}}\\{\left(\substack{{2}\\{\substack{{\vdots}\\{\binom{n-1}{n}}}} }\right)}}\right)$ for every $n\in\mathbb{N}$, then the answer is that the limit does not exist. It is easy to see that $a_n=a_{n+3}$ for all $n\in\mathbb{N}$. With $a_1=1$, $a_2=0$, and $a_3=1$, we conclude that sequence $\left\{a_n\right\}_{n\in\mathbb{N}}$ is a nonconstant periodic sequence, whence the limit $\displaystyle\lim_{n\to\infty}\,a_n$ cannot exist.

Batominovski
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7

What do you mean by "this number"? One reasonable interpretation might be that you're hoping for a limit of a sequence of finitely-nested expressions $a_n$: $$ 1, {1 \choose 2}, {1 \choose {2 \choose 3}}, {1 \choose {2 \choose {3 \choose 4}}}\ldots$$ Now ${a \choose b} = 0$ if $b > a$, while ${a \choose 0} = 1$ and ${a \choose 1} = a$. Thus this sequence will cycle through $1,0,1,1,0,1,\ldots$ and there will be no limit.

Robert Israel
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2

Note that ${i\choose i+1} = 0$.

So $A_2 ={1 \choose 2} = 0$,

$A_3 ={1\choose{2\choose 3}} = {1 \choose 0} = 1$

Now $A_4={1\choose{2\choose {3\choose 4}}} = {1\choose{2\choose 0}} = {1\choose{1}} = 1$

Lastly $$A_5={1\choose{2\choose {3\choose {4\choose 5}}}} = {1\choose{2\choose {3 \choose 0}}} = {1\choose{2\choose 1}} = {1\choose 2} = 0$$

If you consider a translation $${1+1\choose{2+1\choose {3+1\choose {4+1\choose 5+1}}}} = {1+1\choose{2+1\choose {3+1 \choose 0}}} = {1+1\choose{2+1\choose 1}} = {1+1\choose 2+1} = 0$$

So you find a period for your expression.

Consider $$A_n = {1 \choose {2 \choose \underset{n}{\vdots} }}$$

Since $${k+1\choose{k+2\choose {k+3\choose {k+4\choose k+5}}}} = \ldots = 0 = {k+1\choose k+2}$$

We obtain $A_{10} = A_7 = A_4 = 1 $. To the general case take $n \mod 3 = r$ $A_n = A_{r + 3}$ or you could consider $A_0 = 1A_1 = 1$ then $$A_0 = A_3,A_1 = A_4, A_2= A_5 $$

That is $A_n = A_r$ (where $r = n \mod 3$)

Conrado Costa
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