[seqfan] Re: A triangular observation (nothing proved)
Alonso Del Arte
alonso.delarte at gmail.com
Thu Mar 4 14:59:39 CET 2010
For what it's worth and speaking only for myself, I do find your triangle
interesting. But if I had come up with it, I would first investigate any
possible connections to Pascal's triangle before submitting it: Can this
triangle be derived from Pascal's triangle somehow (as can Lozanic's)? If
you subtract this triangle from Pascal's triangle, what is the resulting
triangle? And of course I would also look up the various sequences derived
from the triangle (such as row sums) in the OEIS and see if any interesting
angle emerges.
Al
On Thu, Mar 4, 2010 at 7:29 AM, Jonas Wallgren <jwc at ida.liu.se> wrote:
> Hi!
>
> This is (the top of) a triangle for r=3.
> The left edge is n, the right edge is n². (n is the row number.)
>
> 0
> 1 1
> 2 0 4
> 3 0 0 9
> 4 2 2 2 16
> 5 2 6 6 4 25
> 6 3 3 12 9 9 36
> 7 6 8 10 22 20 18 49
> 8 7 13 19 18 41 35 29 64
>
> Every element inside the triangle is the sum of r+1 elements r rows up.
> E.g.
> 18 in the middle of the bottom row = 2+6+6+4 from 3 rows up. The elements
> summed are those placed symmetrically around a vertical line through the
> sum
> and as close to it as possible.
> All elements outside the triangle = 0.
>
> The sum of all elements in each row gives the following sequence, together
> with its 1st and 2nd differences:
>
> 0 2 6 12 26 48 78 140 234 360 614 996 1506 2528 4062
> 2 4 6 14 22 30 62 94 126 254 382 510 1022 1534
> 2 2 8 8 8 32 32 32 128 128 128 512 512
>
> For an arbitrary r the 2nd difference (seems to) consists of:
> r-1 copies of 2,
> r copies of 2*(r+1),
> r copies of 2*(r+1)²,
> r copies of 2*(r+1)³,
> ...
>
> Maybe this is interesting in some way ... ?
>
> /Jonas
> Wallgren
>
>
>
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