[seqfan] Peculiarity of A130595, Inverse Pascal Triangle

[Dale Gerdemann]

Hello Seqfans,

In the usual Pascal recursion

   T(n,k) = T(n-1,k-1) + T(n-1.k)

each entry in the triangle is dependent on two neighbors above.

But in A130595, you can generate an entry given only one entry above:

   T(n,k) = (n+1)/(k-n-1) * T(n-1,k)

Try for example T(8,5). First locate T(8,5) in the triangle:

Triangle begins:

1;

-1, 1;

1, -2, 1;

-1, 3, -3, 1;

1, -4, 6, -4, 1;

-1, 5, -10, 10, -5, 1;

1, -6, 15, -20, 15, -6, 1;

-1, 7, -21, 35, -35, 21, -7, 1;

1, -8, 28, -56, 70, -56, 28, -8, 1;

-1, 9, -36, 84, -126, 126, -84, 36, -9, 1


So T(8,5) = -56, and -56 * 9/-4 = 126, just as you can see in the triangle.


Has this been discussed anywhere?


Dale