Request for help in extending floretion sequence.

[Creighton Dement]

Dear Seqfans, 

I would like to request help in extending this sequence: 
http://www.crowdog.de/fav.txt

1, 1, 0, 0, 0, 0, -1, 0, 2, 3, 0, -3, 0, 2, 0, 0, 4, 2, -4, -4, 3, 5,
-1, -4, 3, 5, -5, -5, 6, 8, -6, -7, 4, 5, -3, -3, 4, 4, -4, -4, 5, 7,
-3, -6, 3, 7, -2, -6, 3, 6, -3, -6, 4, 8, -2, -7, 1, 7, 0, -5, 2, 5, -3,
-5, 5, 8, -4, -6, 3, 6, -2, -5, 5, 6, -5, -6, 4, 7, 0, -6, -1, 5, 2, -3,
0, 5, 1, -5, -1, 6, 2, -4, -1, 2, 2, -1, 1, 2, -1, -2, 1, 3, 3, -3, -2,
4, 4, -3, -2, 4, 3, -5, -2, 5, 4, -5, -3, 5, 5, -4,  

from currently 30000 terms to as many terms as possible (i.e. as long as
the pattern still looks interesting) - along with a few others like it. 

A plot of the first 30000 terms: http://www.crowdog.de/best30000.jpg


It took my 1800 MHz computer around 2 hrs. to calculate the first 30000
terms of the above sequence (along with 5 of its "neighbors"). 
Basically, I need someone with a faster computer (preferably Unix/Linux
compatible)  and / or more time- I would rather not leave my old
computer running over night. The computer progam FAMP has been
programmed to output it so it is not absolutely necessary to understand
its definition in all detail.  

A sketch of how to compute the above sequence:

If X is any floretion, let X[0] stand for the real coefficient of 'i in
X, X[1] for 'j, X[2] for 'k, X[3] for i', , X[4] for j', X[5] for k',
X[6] for 'ii', X[7] for 'jj', X[8] for 'kk', X[9] for 'ij', X[10] for
'ik', X[11] for 'ji', X[12] for 'jk', X[13] for 'ki', X[14] for 'kj' and
X[15] for the unit e. FAMP represents a floretion X by first calling the
array: double X[] = new double[16] and then assigning the array values. 

Note: (int)x, below, is equivilant to floor[x]. Ex. (int)12.75 = 12. 

Define the floretion: H =  + 'i + .25'ii' + .25'jj' + .25'kk' + .25e

Define the boolean variables wasi, wasj, wask. Initially set all equal
true. 
Set Y = H. 

1. Calculate Y = Y*H 

2. Define emsumi, emsumj, emsumk and sum as:

          emsumi = Y[0] - (int) Y[0] + Y[3] - (int) Y[3] + 

			    Y[6] - (int) Y[6] + Y[7] - (int) Y[7]   + 

			    Y[8] - (int) Y[8] + Y[12] - (int) Y[12]+ 

			    Y[14] - (int) Y[14] + Y[15] - (int) Y[15];

			

	  emsumj = Y[1] - (int) Y[1] + Y[4] - (int) Y[4] + 

			    Y[6] - (int) Y[6] + Y[7] - (int) Y[7]   + 

			    Y[8] - (int) Y[8] + Y[10] - (int) Y[10]+ 

			    Y[13] - (int) Y[13] + Y[15] - (int) Y[15];

				

	  emsumk = Y[2] - (int) Y[2] + Y[5] - (int) Y[5] + 

			    Y[6] - (int) Y[6] + Y[7] - (int) Y[7]   + 

			    Y[8] - (int) Y[8] + Y[9] - (int) Y[9]+ 

			    Y[11] - (int) Y[11] + Y[15] - (int) Y[15];

	    sum = emsumi + emsumj + emsumk; 
	

3. Run through the following conditions:
  
	if (emsumi > emsumj && emsumi > emsumk) {

							if (wasj)

								Y[0] = Y[0] - emsumi;

							else

								Y[0] = Y[0] + emsumj + emsumk;

							wasi = true;

							wasj = false;

							wask = false;

						}



						else if (emsumj > emsumi && emsumj > emsumk) {

							if (wask)

								Y[1] = Y[1] - emsumi;

							else

								Y[1] = Y[1] + emsumi + emsumk;

							wasj = true;

							wasi = false;

							wask = false;

						}



						else if (emsumk > emsumi && emsumk > emsumj) {

							if (wasi)

								Y[2] = Y[2] - emsumi;

							else

								Y[2] = Y[2] + emsumj + emsumi;

							wask = true;

							wasi = false;

							wasj = false;

						}



						else {



							Y[15] = Y[15] + sum;

							wasi = true;

							wasj = true;

							wask = true;



						}
						
Let a(0) be 2*(Y[0] + Y[1] + Y[2] + Y[3] + Y[4] + Y[5] + Y[6])

Return to step 1. At the end of step 2, let a(1) be 2*(Y[0] + Y[1] +
Y[2] + Y[3] + Y[4] + Y[5] + Y[6]), etc. 


Sincerely, 
Creighton 
						
-It's a shame when the girl of your dreams would still rather be with
someone else when you're actually in a dream.