Re: referencesfor these coefficients



Am 10.09.05 01:11 schrieb Gottfried Helms:
....

To give a short example:

Add the products of the coefficients of the Eulerian triangle with the
appropriate number of consecutive binomial-coefficients of the list
right to it to get the powers of n (rhs)
>
>>Here's a triangle which I find somewhat interesting. The first few rows are:
>>
>>
>> 1
>> 1 1 * 0 0 0 1 3 6 10 15 21 ... = 1 4 9 16
>> 1 4 1 * 0 0 0 1 4 10 20 ... = 1 8 27 64
>> 1 11 11 1 * 0 0 0 1 5 15 35 ... = 1 16 81 256 ...
>> 1 26 66 26 1
>> 1 57 302 302 57 1
>> 1 120 1191 2416 1191 120 1
>>
If you shift the binomial-rows one row up, you get the sum-of-powers
(first row lhs: sum of n = powers n^1, second row sum of squares, third row sum of cubes...)

>> 1 * 0 0 0 1 3 6 10 15 21 ... = 1 3 6 10
>> 1 1 * 0 0 0 1 4 10 20 ... = 1 5 14 30
>> 1 4 1 * 0 0 0 1 5 15 35 ... = 1 9 36 100 ...
>> 1 11 11 1
>> 1 26 66 26 1
>> 1 57 302 302 57 1
>> 1 120 1191 2416 1191 120 1

>From that you can extend this to negative rowindices (and simply determine the question-marks)

>> 1 ? ? * 0 0 1 -3 3 -1 0 0 = 1 1/16 1/81 1/256
>> 1 ? ? * 0 0 1 -2 1 0 0 0 = 1 1/8 1/27 1/64
>> 1 ? ? * 0 0 1 -1 0 0 0 0 = 1 1/4 1/9 1/16
>> 1 1/2 1/3 * 0 0 1 0 0 0 0 0 = 1 1/2 1/3 1/4
>> 1 0 * 0 0 0 1 1 1 1 1 1 = 1 1 1 1
>> 1 * 0 0 0 1 2 3 4 5 6 = 1 2 3 4
>> 1 1 * 0 0 0 1 3 6 10 15 21 ... = 1 4 9 16
>> 1 4 1 * 0 0 0 1 4 10 20 ... = 1 8 27 64
>> 1 11 11 1 * 0 0 0 1 5 15 35 ... = 1 16 81 256 ...
>> 1 26 66 26 1
>> 1 57 302 302 57 1
>> 1 120 1191 2416 1191 120 1

If I recall it right, then even the interpolation between rows was
meaningful in this system, where only the negative-indexed rows were
of little use, since the sequence of the binomial-array get finite
then and thus the sequences of the rows of the eulerian "triangle"
get infinite. But just for curiousity I would try to see, what this
would be looking like.

Gottfried Helms


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