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 |