# Is this variation and its characteristic of binomial coefficient known or is it just another terrible deterministic integer factorization algorithm? [closed]

I added +341(-341 is also good) to the binomial coefficient function and came up with this.

and residues of mod N (vertical line-1)

Gerry. Yes numbers are misrepresented but they do grow. The problem is that boxes are not resizible. line 0 is 1; line 1 is 1,1; line 2 is 1,343(2+341),1; line 3 is 1, 685(1+343+341),685,1; line 4 is 1, 1027 (1+685+341), 1717(2*685+341),1027,1;

line 11 gives numbers that are divisible by 11 and line 31 gives numbers that are divisible by 31

They are factors of 341.

-
or residues of diagonals i.stack.imgur.com/vNmNo.png –  Bojan Vasiljević Mar 11 '12 at 22:34
The images make no sense to me. If all you do is add 341 to the binomial coefficients they still get bigger from one row to the next, while in your first table they sometimes get smaller. I don't see why certain rows are highlighted, and I don't see what any of it has to do with factorization. You have a lot of work to do to make this a real question. –  Gerry Myerson Mar 11 '12 at 23:30
You may know that if $p$ is prime then all the numbers $p$-choose-$k$ are multiples of $p$ (except for $k=0$ and $k=p$). In particular, all the inner entries in row 11 of Pascal's triangle are divisible by 11, and all the entries in row 31 are divisible by 31. Adding a multiple of $341=11\times31$ won't affect this divisibility property for those two rows.