Actually, I do... it all started way back at Kutztown University. I was a math major. Then I realized that computers were more my thing. One day I was screwing around with some numbers and came upon this little formula (actually, I just found the pattern). Four years later, Dr. Thang Bui from Penn State Harrisburg helped me get the formula into an explainable format.

Basically, what the formula says is this:

- take an arbitrary n
- take n+1 (k) positive consecutive integers
- take those integers to the power n
- subtract the lesser from the greater going down the column, forming a new column
- this should leave you with a column containing one less element than before
- keep doing this until you get one number (a total of n columns of subtraction)
- this number is n factorial (n!)

k^3

----

1^3 = 1 | |||

7 | |||

2^3 = 8 | 12 | ||

19 | 6 | ||

3^3 = 27 | 18 | ||

31 | |||

4^3 = 64 |

if you were to continue this by adding 5^3 to the leftmost column and began subtracting in the same way, the answer after n columns of subtraction would still be 6. you can start at 1456 and the same thing would happen (1456, 1457, 1458, 1459).

this is also handy for showing that 0! = 1 (although NOT a proof!)

pick n = 0 and n + 1 arbitrary postive consecutive integers (in this case, 17, thanks Dr. Hartzler)

k^0

----

17 ^ 0 = 1 |

Here's the equation for this bad boy:

I've written a program to replicate this and show that it works but i don't have it here yet. It was written for Linux (which you should be using anyway). If i recall correctly, the program works until 1178, but the segfaults get all pissy after that. An issue i will try to resolve. And it takes a loooonnnggg time to compute 1178 (1179^1178 is one big-ass number). So plan on this thing sitting around for a little while. I won't be releasing the source code because it's a foul piece of shit. Global arrays of char's flying everywhere, and i HATE commenting. If it's tough to write, it should be tough to read.

Stay tuned for my next formula! It discovers another pattern in Pascal's Triangle...

if you read each row of pascal's triangle as one concatenated number, that number is 11^(row index base 0). so 1 = 11^0, 11 = 11^1, 121 = 11^2, and so on. BUT... it stops on the sixth line... 1 5 10 10 5 1 (or 15101051, concatenated) is NOT 11^5! i can show you that it IS with an algorithm i've created. This one was created AFTER i left kutztown so don't go thinking that kutztown is some kind of a mecca for creative math people. More later... (although flooding my mailbox will cause me to get the formula up quicker)

Any questions or comments? Email Me

Meanwhile, here's a little something I fired up in Maple V. Bitchin':

**f(x,y) = sin(x)cos(y), g(x,y) = sin(y)cos(x)**