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## Singmaster’s Conjecture

The Singmaster conjecture, named after David Singmaster, is about how many times different numbers appear in Pascal’s triangle. It says to display any number of times at most, except one.

Clearly, the number 1 occurs an infinite number of times, since every row begins and ends with 1. But the second diagonal of Pascal’s triangle is counting numbers 2,3,4,5… , and whatever When you choose a number, there will be a row that is the second and last number. After this line, all numbers except 1 are greater than the number you selected. Therefore, we can conclude that every number appears a finite number of times in Pascal’s triangle, except for 1.

Singmaster asked next: Is there a limit to the number of repetitions that cannot be exceeded *whatever *number? For example, can we say that whatever number you choose, let it be so large that it never appears more than, say, 20 times in Pascal’s triangle? Here are some examples of how many times different numbers appear in Pascal’s triangle:

1 appears an infinite number of 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 appear six times each

3003 appears eight times

These examples show that there are surprisingly few repetitions of numbers in Pascal’s triangle, with the vast majority of numbers occurring only twice (as the second and second last digit in a row). Additionally, no other number has been found that occurs more than 6 times besides 3003, and no numbers have been found that occur five or seven times, but no one knows for sure if such numbers exist.

But before you turn away in disgust at what useless and lazy schemes mathematicians are, Singmaster found something interesting about six or more occurrences – he proved that there are infinitely many. In fact, he found a formula that always gives you numbers like this:

n = F(2i+2)*F(2i+3)-1

k = F(2i)F(2i+3) +1

Here F(I) represents the Ith Fibonacci number (numbers from the sequence 1,1,2,3,5,8,13,21,34,55… where each number is the sum of the previous two) . After calculating n and k, you need to calculate n choose k = n!/(r!(nr)!) to get the real number that appears six or more times. Thus, if we choose, for example, I = 1, 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

and k = F(2*1)F(2*1+3) +1

= F(2)F(5) + 1

= 1*5 +1

= 6

Finally, we calculate 14 and select 6:

14!/(6!(14-6)!)

= 87178291200/(720*40320)

= 87178291200/(29030400)

= 3003

So the first number in the 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 repetitions, but *6 or more.* Even though 3003 actually appears 8 times, the formula still holds. Second, the Singmaster formula doesn’t deliver *age *a 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 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 an infinite number of numbers appearing 6 or more times.

By the way, the next number you get from the Singmaster formula (using I = 2) is 61218182743304701891431482520.

I find it quite fascinating how something as simple as Pascal’s triangle, which is ultimately just a series of very simple additions, can produce puzzles like this. It is quite humbling to think that such simple questions, so easy to think of, have answers that continue to confound us as their answers remain elusive.

If you are interested in learning more about Pascal’s Triangle and its amazing patterns, properties and secrets, including its connection to the Fibonacci numbers (which here seemed to just pop out of thin air), please visit my site by following the link in the resource box below.

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