# Does this recursive journey through Pascal's triangle always reach $1$?

Let $$p(k)$$ be the $$k^\text{th}$$ number in Pascal's triangle, numbered from left to right in each row, going down the rows.

So, for example, $$p(1)$$ to $$p(10)$$ are $$\binom{0}{0},\space \binom{1}{0},\space \binom{1}{1},\space \binom{2}{0},\space \binom{2}{1},\space \binom{2}{2},\space \binom{3}{0},\space \binom{3}{1},\space \binom{3}{2},\space \binom{3}{3}$$, respectively.

Now define sequence $$a_n$$ as:

$$a_1\text{ is a positive integer}$$

$$a_{n+1}=p(a_n)$$

Is the following conjecture true or false:

Conjecture: $$\lim\limits_{n\to\infty}a_n=1$$ for all $$a_1\in\mathbb{Z^+}$$.

For example, with $$a_1=150$$ we have:
$$a_2=p(150)=560$$
$$a_3=p(560)=32$$
$$a_4=p(32)=35$$
$$a_5=p(35)=7$$
$$a_6=p(7)=1$$
$$a_n=1$$ for $$n\ge 6$$

With $$a_1=100$$ we have:
$$a_2=p(100)=1287$$
$$a_3=p(1287)=37353738800$$
$$a_4=p(37353738800)=\space ?$$

If an $$a_n$$ value is very large, then $$a_{n+1}$$ is likely to be much larger than $$a_n$$. However, even if the terms become very large, it is possible that eventually one of them will be small enough so that the sequence will become an endless stream of $$1$$s. I don't know how the sequence will play out.

Here and here is the sequence $$p(k)$$, except their index is shifted down $$1$$ from mine. So for example, my $$p(5)$$ is numbered as the $$4$$th term is the linked sequence.

Context: I have been playing with Pascal's triangle, investigating some of its mysterious, geometrical and humourous properties.

• The first step to attack this problem is to find a way to efficiently determine $p(n)$ for very large $n$ Dec 8, 2023 at 6:54
• I don't know about the efficient part but at least $p(n)=\binom{r-1}{n-1-\frac{r(r-1)}{2}}$ where $r=\lfloor \frac{1+\sqrt{8n-7}}{2} \rfloor$
– Sil
Dec 8, 2023 at 16:05

At least heuristically, there should be an $$a_1$$ for which the $$a_n$$ tend to infinity. First of all, we can take $$a_1$$ large enough to consider only asymptotics. Now, note that $$p(k)$$ 'looks like' $${\lfloor c\sqrt{k}\rfloor\choose i}$$ for some constant $$c$$ and an $$i$$ between $$0$$ and $$\lfloor c\sqrt{k}\rfloor$$. We can (heuristically) assume that this $$i$$ is uniformly distributed over its range. Then if $$8\leq i\leq \lfloor c\sqrt{k}\rfloor-8$$, then $$p(k)\geq C_1(c\sqrt{k})^8 \geq C_2 k^4$$ for some $$C_1$$ and $$C_2$$. In other words, with probability $$1-\frac{C_3}{\sqrt{a_n}}$$ we have $$a_{n+1}\geq C_2{a_n}^4$$. But then with probability $$1-\frac{C_3}{\sqrt{a_{n+1}}}$$ we have $$a_{n+2}\geq C_2a_{n+1}^4$$. So the probability that this always happens is $$\left(1-C_3(a_1)^{-1/2}\right)\left(1-C_3(C_2a_1^{-2})\right)\left(1-C_3(C_2(C_2^4a_1^{-8}))\right)\cdots$$, and by the standard convergence theorems for products this converges to a value $$\gt0$$.