I was recuperating last year from my heart attack and got very bored with the TV, so I had my friend Terry bring me a calculator so I could have something interesting to do.

Well, for a couple of weeks or so I calculated all of the prime factors of the composite numbers up to 5000, after which I was again bored.  So I decided to see if I could find a way to generate the diagonals of Pascal's Triangle. I didn't really expect to succeed,  but I figured it would be fun to give it a try.  Well, I was right, it was a lot of fun! Especially when I actually succeeded!

I don't want to die with this thing stuck in my head,  so I want to get it out into the world so it won't disappear when I do.  I'm bringing it here first just because someone here may be able to tell me if someone has come up with it before or if I'm the first to discover it. I've been searching the internet for a long time, but I've been unable to find it anywhere. I know how to generate Pascal's Triangle row by row using addition, but I just wanted to see if I could do it this way.

The first two diagonals are pretty trivial, since the first is composed entirely of 1s, and the second is composed of the counting numbers, neither of which needs an algorithm to be generated, though my method works for them as well as all of the others.

The third diagonal is a bit more interesting, being composed of the triangular numbers.  So let's begin there:

This diagonal starts with a 1 and a 3. Start by multiplying
1x3/1 to get a 3.
Next multiply the 3 by 4/2 to get 6.
Next multiply the 6 by 5/3 to get 10.
Next multiply the 10 by 6/4 to get 15.
Next multiply the 15 by 7/5 to get 21.

Continue in this way as long as you like to get as many of the triangular numbers as you want, each time multiplying the result by the next fraction, each time adding 1 to both the numerator and the denominator to get the next number in the sequence.

The fourth diagonal is composed of the tetrahedral numbers, and can be generated thusly:

Begin by multiplying 1 by 4/1 to get 4.
Next multiply 4 by 5/2 to get 10.
Next multiply 10 by 6/3 to get 20.
Next multiply 20 by 7/4 to get 35.
Next multiply 35 by 8/5 to get 56.

Continue as long as you like, each time again adding 1 to both the numerator and the denominator to get the next number.

See how simple and easy this is? Every diagonal can be generated this way, though I don't know how to prove it rigorously, not being a mathematician.

Let's do one more random diagonal just to see if it continues to work:

Let's go with the diagonal that begins with a 1 and a 17:

1x17/1 = 17
17×18/2 = 153
153×19/3 = 969
969×20/4 = 4845
4845×21/5 = 20349
20349×22/6 = 74613

And I could continue this until the numbers get too big for my calculator.

Well, what do you think? Have I come up with something new or am I full of shit?
I think it's a really beautiful algorithm, but I've never had any feedback about it from anyone who knows or cares about recreational math. I'm thinking of emailing it to Mathologer (Burkard Polster) of YouTube fame, but I'm afraid he'll just ignore me, since I'm a nobody.

I am not a smart man. Especially in regards to this kind of stuff.

Just want to say, mail it to the youtube smart guy. Worst case, It'll be ignored. You'll lose nothing.

And if he tries to steal credit, you have the mail as evidence that it is your Discovery.

Best case, you get proven right and you get Lauded.

Nothing to lose, anything to gain, I think.

What do I think???  I think you are so far advanced in this stuff that I could never (if even then!) understand what you just did!! 

Quote from: Mr.Obvious on July 28, 2023, 08:43:38 AMI am not a smart man. Especially in regards to this kind of stuff.

Just want to say, mail it to the youtube smart guy. Worst case, It'll be ignored. You'll lose nothing.

And if he tries to steal credit, you have the mail as evidence that it is your Discovery.

Best case, you get proven right and you get Lauded.

Nothing to lose, anything to gain, I think.

Hell, I really don't care about credit,  I just want to know if it's been done before and whether a real mathematician finds it as beautiful and elegant as do I. It would be pretty cool if Mathologer were to make one of those cool videos about it like he's done about so many other things in math. If he does ignore me I'll give it to Numberphile or the guy who does Stand Up Maths. 🖖

Maybe the answer you seek is here-

wiki

I just discovered today another way of generating the sequence of numbers in the diagonals of Pascal's Triangle!
I use a calculator that has a "table" button, it makes it a lot easier.  It's a TI-30XS MultiView. The table function takes an expression with 1 variable (x) and returns a list of all of the associated "y"s.

This method uses the "choose" function,  n choose k:

x nCr 0 returns a string of 1s, the first diagonal of the Triangle.

x nCr 1 gives the sequence of positive integers.

x nCr 2 gives the triangular numbers.

x nCr 3 gives the tetrahedral numbers.

x nCr 4 gives the 5-simplex numbers.

x nCr 5 gives the 6-simplex numbers.

And so on, all the way down the Triangle.

Maybe someone already knows about this, in which case I bow to them. But then again, maybe no one else knows this, in which case I bow to myself!

🥸