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Each number occurs at least twice as $a={a\choose1}={a\choose a-1}$. If a number occurs somewhere else in the triangle (most likely twice, if it's not of the form ${2a\choose a}$) then that number occurs $4$ times. After that, it becomes interesting. I found the following with a simple script:

$$120={120\choose1}={16\choose 2}={10\choose3}={10\choose7}={16\choose 14}={120\choose119}$$

$$210={210\choose1}={21\choose 2}={10\choose4}={10\choose6}={21\choose 19}={210\choose209}$$

$$1540={1540\choose1}={56\choose 2}={22\choose3}={22\choose19}={56\choose 54}={1540\choose1539}$$

$$7140={7140\choose1}={120\choose 2}={36\choose3}={36\choose33}={120\choose 118}={7140\choose7139}$$

And a special one, that did not only occur six times, but eight:

$$3003={3003\choose1}={78\choose 2}={15\choose5}={14\choose6}={14\choose8}={15\choose10}={78\choose 76}={3003\choose3002}$$

I only checked the numbers up to $10000$; so here's my question

Besides $1$, are there other numbers that occur infinitely often? Is there an upper bound known to how many times a number can occur in Pascal's Triangle?

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Singmaster's conjecture (Wikipedia article) is that there is a finite upper bound on the multiplicities of entries in Pascal's triangle (other than 1). The article has all of the latest developments in this direction, and as explained in the article,

It is clear that the only number that appears infinitely many times in Pascal's triangle is $1$, because any other number $x$ can appear only within the first $x + 1$ rows of the triangle.

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    $\begingroup$ I feel silly for asking whether numbers can occur infinitely often now, that was pretty obvious. Arbitrarily many times, however, remains interesting. Thanks! $\endgroup$ – vrugtehagel Apr 2 '17 at 18:29

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