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

[Bob Selcoe]

Hi Dale and Seqfans,

Unless I'm missing something, this holds also for Pascal's Triangle, because 
C(n,k) = n!/k!*(n-k)!  So C(n+1, k+1) = (n+1)!/(k+1)!*(n-k+1)!, which is 
another way of presenting your observation.

Cheers,
Bob Selcoe
--------------------------------------------------
From: "Dale Gerdemann" <dale.gerdemann at gmail.com>
Sent: Monday, January 05, 2015 12:49 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
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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