[seqfan] Re: Peculiarity of A130595, Inverse Pascal Triangle

[Frank Adams-Watters]

One can make the same kind of calculation for the Pascal triangle; just 
replace the (k-n-1) with (n-k+1). Or if you prefer,

C(n,k) = n/k * C(n-1,k-1)

This is not a difficult result; it follows immediately from C(n,k) = 
n!/(k! * (n-k)!).

Franklin T. Adams-Watters

-----Original Message-----
From: Dale Gerdemann <dale.gerdemann at gmail.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Mon, Jan 5, 2015 2:04 pm
Subject: [seqfan] Peculiarity of A130595, Inverse Pascal Triangle


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

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