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Henry

Newbie Joined: 02 Jun 2010
Posts: 7 Posted: 05 Jun 2010 03:47:23 pm    Post subject: Ever think about figurative numbers? Triangular numbers follow the series 1,3,6,10,15,..., which is the third left-going diagonal in Pascal's triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Also there are square, pentagon,... numbers which follow the third left-going diagonal of Lucas Triangles, with a different initial G(1) = d.*explained in next paragraph. 1,4,9,16,25,... 1,5,12,22,35,... . . . These numbers can be explained by the cumulative sums of an initial arithmetic series, where the difference "d" is equal to the sides of the figurative number series of interest minus two. Thus triangular numbers (sides = 3), would have a "d" of 1. The triangular numbers are then seen as the cumulative sum of the series 1,2,3,4,5,... Same goes for the other figurative numbers, but with a different initial series, with a different "d" Using a Binomial Transform on the different series one can find the common equation pattern to be n(dn-(d-2))/2, where "n" is the nth term of the figurative number sequence of interest. But, mix it up a bit: what if you want figurative numbers in different dimensions? One could imagine this in triangular form (looking down) graphically as: (third dimension) this picture: And one could then imagine this for different d's or different polygon number series. The fourth dimension, however is harder to imagine, and higher even harder to imagine. But what is the mathematical connection? Well, each dimension is a corresponding diagonal on each corresponding triangle figure with G(0) = 1 and G(1) = d. So with huge amounts of simplification, one can arrive at the conclusion that a figurative number, n, of d, of dimension z is this program (whose equation I don't want to write out in text, because it would be ugly on text) Code: ``` ClrHome Disp "    σ(n,d,z) Input " n: ",N Input " d: ",D Input " z: ",Z If N+Z>1 Then (DN-D+Z)(N+Z-2)!/Z!/(N-1)!→A Else D→A End Output(6,1,"σ(n,d,z) = Output(6,13,A```Last edited by Guest on 07 Jun 2010 02:09:43 pm; edited 1 time in total
Weregoose
Authentic INTJ

Super Elite (Last Title) Joined: 25 Nov 2004
Posts: 3976 Posted: 05 Jun 2010 09:48:53 pm    Post subject: Very cool. It's not easy to find text that goes so deep. Linear recurrences were absolutely a fun study for me. The calculator's Seq mode is a great way to explore their beauty, but as far as practicality goes, many of its results can be boiled down to simple expressions, anyhow. Dividing (DN-D+Z)/Z! out of your line, I can see that the remainder may be replaced by (N+Z-2) nPr (Z-1).Last edited by Guest on 05 Jun 2010 09:50:32 pm; edited 1 time in total
Henry

Newbie Joined: 02 Jun 2010
Posts: 7 Posted: 06 Jun 2010 11:36:19 pm    Post subject: Cool trick, I'll see if it works out. Right now I'm in the works with trying to figure out a universal formula for a zth demensional sum (or just how many sums of sums) of an initial sequence f(x), so that in the zth dimension, the sum would be g(x). - unlike this case, where the initial sequence f(x) is always arithmetic, in the wicked simple form y=mx, where m=d. And like you said, delta seq can do a lot to help with programing these types of things. I actually just discovered delta seq not that long ago. One can do binomial transforms, generating functions, Euler transform in a jiff! Which means one could emulate a sequence pretty accurately with the transform function. Hey do you think that with data, a transform could be as accurate, or not as some regression of whatever degree function? I will have to test that out.
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