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.

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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:

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


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:
$a_n=1$ for $n\ge 6$

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

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.

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

1 Answer 1


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 $\gt0$.


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