The Diagonals of Pascal's Triangle

I'm looking for some feedback about 2 methods I've found for generating the sequences of numbers in the diagonals of Pascal's Triangle. The first one is something I came up with about a year ago, while ensconced in a nursing facility recuperating from a heart attack. Once I'd read all of the books they had that were of any interest, I quickly grew bored with the TV, so I asked a friend of mine to bring me a calculator so I could play with numbers, which I sometimes enjoy doing. I am not a mathematician, but I sometimes enjoy playing with numbers the way children play with toys.

I spent a couple of weeks finding the prime numbers and the prime factors of the composite numbers up to 5,000, which was about as far as I wanted to go with that particular project.

So next I turned my attention to Pascal's Triangle, which I had enjoyed reading about before, since it's among the most interesting number patterns I've come across.

I knew how to build the Triangle row by row using addition, and I knew it had many interesting patterns within it. I spent some time reacquainting myself with those, and then decided to see if I could find a method of generating the sequences of numbers in the diagonals of the Triangle.
I didn't really expect to succeed, but I knew that the only things worth doing are the hard things.

So I got busy, and I was right about the difficulty - it was a bit hard.

But then I actually succeeded!

Let me show you what I found:

The first two diagonals are pretty trivial, and don't need an algorithm to generate them, so I'll begin with the third diagonal, which is composed of the triangular numbers:

1×3/1=3
3×4/2=6
6×5/3=10
10×6/4=15
15×7/5=21
21×8/6=28
28×9/7=36
36×10/8=45
45×11/9=55
55×12/10=66
66×13/11=78
78×14/12=91
Etc.

Next, let's look at the diagonal that contains the tetrahedral numbers:

1×4/1=4
4×5/2=10
10×6/3=20
20×7/4=35
35×8/5=56
56×9/6=84
84×10/7=120
120×11/8=165
165×12/9=220
220×13/10=286
286×14/11=364
364×15/12=455
Etc.

The next diagonal consists of the 5-simplex numbers:

1×5/1=5
5×6/2=15
15×7/3=35
35×8/4=70
70×9/5=126
126×10/6=210
210×11/7=330
330×12/8=495
495×13/9=715
715×14/10=1001
1001×15/11=1365
1365×16/12=1820
Etc.

OK, I'll just do one more, because I expect that by now you can see the basic idea:

This diagonal is composed of the 6-simplex numbers:

1×6/1=6
6×7/2=21
21×8/3=56
56×9/4=126
126×10/5=252
252×11/6=462
462×12/7=792
792×13/8=1287
1287×14/9=2002
2002×15/10=3003
3003×16/11=4368
4368×17/12=6188
Etc.

See how simple and easy that is?
I could go on, but I don't want to bore y'all to tears, so I'll stop here.

I came up with that about a year ago, but just last week I found a completely different method for doing the exact same thing!

This one uses combinations, which are pretty easy to do with the calculator I have, a TI-30XS MultiView. It has a "table" button, which allows me to input an expression with one variable (x), and it returns a list of the appropriate "y" values.

The first diagonal, composed of a potentially infinite sequence of 1s, is arrived at this way:

x nCr 0

The next diagonal, composed of the positive integers, is arrived at with

x nCr 1

Next, the triangular numbers:

x nCr 2

Then, the tetrahedral numbers:

x nCr 3

Then the 5-simplex numbers:

x nCr 4

Then the 6-simplex numbers:

x nCr 5

Etc.

This works for all of the diagonals of the Triangle.

I've been searching on Google and YouTube for any hint that someone else has come up with these two methods, but I haven't been able to find anything like them.

So I'm hoping that someone here can help me find out whether or not anyone has done either of these methods before or if I'm the first to discover them.

Also, are they as beautiful as I, a non-mathematician, think they are, or are they just too trivial to care about?

I really need some feedback on this!

Thanks for your time!

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