The Singmaster conjecture, named after David Singmaster, deals with how many different numbers appear in Pascal’s triangle. States that, other than one, there is a maximum number of times any number will appear.
It is clear that the number 1 will appear infinitely many times, since each row begins and ends with a 1. However, the second diagonal in Pascal’s triangle is the counting numbers 2,3,4,5… , and so on. whatever number you choose, comes a row where it is the second and next to last number. After this row, all the numbers except 1 are going to be greater than the number you chose. Therefore, we can conclude that every number appears a finite number of times in Pascal’s triangle except 1.
What Singmaster went on to ask is this: is there a finite number of repetitions that cannot be exceeded by none number? For example, can we say that any number you choose, however large, will never appear in Pascal’s triangle more than, say, 20 times? Here are some examples of how many different numbers appear in Pascal’s triangle:
1 appears infinitely many times
2 is the only number that appears only once
3,4,5,7,8 all appear only twice
6 appears three times
10 appears four times
120, 210, 1540 each appear six times
3003 appears eight times
These examples reveal that there are surprisingly few number repetitions in Pascal’s triangle, with the vast majority of numbers appearing only twice (as the second and penultimate number in a row). Also, no other number has been found to appear more than 6 times except 3003, and no numbers have been found to appear five or seven times, but no one knows for sure if such numbers exist.
However, before you walk away in disgust at how useless and lazy mathematicians are, Singmaster discovered something interesting about numbers that appear six or more times: he proved that there are an infinite number of them. In fact, he found a formula that always gives you those numbers:
n = F(2i+2)*F(2i+3)-1
k = F(2i)F(2i+3) +1
Here F(I) denotes the I Fibonacci number (the numbers in the sequence 1,1,2,3,5,8,13,21,34,55… where each number is the sum of the previous two) . Once you have calculated nyk, to get the real number that occurs six or more times, you must calculate n and choose k = n!/(r!(nr)!). So if we choose I = 1 as an example, we get
n = F(2*1+2)F(2*1+3) -1
= F(4)F(5) – 1
= 3*5 -1 (The fourth and fifth Fibonacci numbers are 3 and 5.)
= 14
yk = F(2*1)F(2*1+3) +1
=F(2)F(5) + 1
= 1*5 +1
= 6
Finally, we calculate 14 we choose 6:
14!/(6!(14-6)!)
=87178291200/(720*40320)
=87178291200/(29030400)
= 3003
So, the first number in Singmaster’s formula that has 6 or more occurrences in Pascal’s triangle is 3003. Two things are worth mentioning here: First, the formula doesn’t give you numbers with exactly 6 occurrences, but 6 or more So even though 3003 actually appears 8 times, the formula still holds true. Second, the Singmaster formula does not give everyone number of six or more digits. Recall from our previous examples that 120, 210, and 1540 are all smaller examples than 3003 that have six or more occurrences in Pascal’s triangle. However, you can replace the “I” in the formula with any integer you like, so the Singmaster formula still shows that there are an infinite number of numbers that occur 6 or more times.
By the way, the next number you get from Singmaster’s formula (using I = 2) is 61218182743304701891431482520.
I find it quite fascinating how something as simple as Pascal’s triangle, which is after all just a series of very simple sums, can give rise to mysteries like this. It’s quite humbling to think that such simple questions that are so easy to conjure up have answers that continue to confuse us with their answers still out of reach.
If you are interested in learning more about Pascal’s triangle and its amazing patterns, properties, and mysteries, including how it relates to the Fibonacci numbers (which here seemed to come out of nowhere), please visit my site by following the link at the resource box below.