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

[Bob Selcoe]

Hi Dale & Seqfans,

I sent this reply earlier today, in response to your first posting - the one 
for which Frank Adams-Watters also replied.     It appears I sent mine right 
after Frank did, but mine got posted on the mailing list later.  We were 
saying basically the same thing.

Cheers,
Bob

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From: "Dale Gerdemann" <dale.gerdemann at gmail.com>
Sent: Monday, January 05, 2015 11:25 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Peculiarity of A130595, Inverse Pascal Triangle

> 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
>
>
>
>
>
>
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>
> 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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>>
>
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