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

[Dale Gerdemann]

Hello Bob, Hello SeqFans,

The identity doesn't hold for Pascal's triangle because::

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

Probably I wasn't clear enough about what f(n) means. My tiling approach
uses different functions for different triangles. For Pascal's triangle,
the function is the constant function: f(n) = 1. Basically what this means
is that you don't need colored tiles for Pascal's triangle. To count T(n,k)
just count the set of length n tilings with k "up" tiles and n-k "down"
tiles. The names "up" and "down" don't help much here; the point is to use
just two different types of tiles.

The identity does hold for A130595 because for this case f(n) = -n. But
it's probably more interesting to look at the other case where the identity
holds, which is A010048 Fibonomal. In this case, f(n) = Fibonacci(n), and
the identity is:

   Fibonacci(n-k)*T(n,k) = Fibonacci(n)*T(n-1,k)

One reason for developing a uniform tiling approach across a wide range of
triangles is so that variant forms of identities can be proven across these
triangles. For example, variant forms of the hockey stick identity can be
proven for the Fibonomial and other triangles by conditioning on the
position of the last up tile or down tile. So I thought that variant forms
of the Inverse Pascal/Fibonomial identity might occur across triangles, but
apparently this is not the case.

Dale















On Mon, Jan 5, 2015 at 10:21 PM, Bob Selcoe <rselcoe at entouchonline.net>
wrote:

> 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
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>>
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