Singmaster’s conjecture, named after David Singmaster, is about how many times different numbers appear in Pascal’s triangle. It states that, apart from one, there is a maximum number of times that any number will appear.
It is clear that the number 1 will appear infinitely many times, as every row begins and ends in a 1. However, the second diagonal in of Pascal’s triangle is the counting numbers 2,3,4,5… , and so whatever number you pick, there comes a row which it is the second and second last number in. After this row, all the numbers except 1 are going to be greater than the number you have chosen. 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 which cannot be exceeded by any number? For example, can we say that whatever number you pick, however large, it will never appear in Pascal’s triangle more than, say 20 times? Here are some examples of how many times different numbers appear in Pascal’s triangle:
1 appears infinitely many times
2 is the only number to appear just 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 repetitions of numbers in Pascal’s triangle, with the vast majority of numbers appearing only twice (as the second and second last number in a row). Furthermore, no other number has been found that appears more that 6 times except 3003, and no numbers have been found that appear five or seven times, but nobody knows for sure if such numbers exist.
However, before you turn away in disgust at what useless, lazy layabouts mathematicians are, Singmaster did find something interesting out about numbers appearing six or more times – he proved that there are infinitely many of them. In fact, he found a formula which always gives you such numbers:
n = F(2i+2)*F(2i+3)-1
k = F(2i)F(2i+3) +1
Here F(I) denotes the I th Fibonacci number (the numbers from 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 n and k, to get the actual number that appears six or more times you have to calculate n choose k = n!/(r!(n-r)!). 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.)
and k = F(2*1)F(2*1+3) +1
= F(2)F(5) + 1
= 1*5 +1
Finally, we calculate 14 choose 6:
So the first number from Singmaster’s formula that has 6 or more occourances in Pascal’s triangle is 3003. It’s worth mentioning two thing here: First, the formula doesn’t give you numbers with exactly 6 repetitions, but 6 or more. So even though 3003 actually appears 8 times, the formula is still valid. Second, Singmaster’s formula does not give every number with six or more digits. Recall from our earlier examples that 120, 210 and 1540 are all smaller examples than 3003 that have six or more appearances in Pascal’s triangle. However, you can replace the “I” in the formula with any integer you like, so Singmaster’s formula still shows there are an infinite number of numbers appearing 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 additions, can give rise to such mysteries as this. It is quite humbling to think that such simple questions that are so easy to conjour up have answers that continue to confound us with their answers still remaining out of reach.
If you are interested in finding out more about Pascal’s triangle and its amazing patterns, properties and mysteries, including how it is linked with Fibonacci numbers (which here seemed to just pop out of thin air), then please visit my site by following the link in the resource box below.