Quaternion "loop" sequence

[Creighton Dement]

Hi all, 

None of the graphed sequences mentioned in my last posts have been
submitted (see http://www.crowdog.de/20801/home.html   ) - mainly
because I first wanted to complete the syllable explanation page
http://www.crowdog.de/Explanations.html
and that is taking me a long time (still unsure about the name field-
perhaps it can be made similar to the one given on this page after the
rules have all been written down). 

In the mean time, I submitted a sequence using only elements from the 
quaternion algebra (over reals) ; the procedure for calculating the
elements of this sequence is very similar to the other "graphed"
sequences.  Perhaps the graph isn't quite as spectacular as the others,
but it seemed simpler for me to explain...  (Maple finds no g.f. after
200 or so terms). I still have 3 more of these to submit but I wanted to
wait to see if anyone has recommendations on things I should change,
first.

 %I A108618
%S A108618  1, 2, -1, -2, -3, -6, -6, 1, 4, 3, 0, -5, -10, -8, 3, 8, 5,
-2, -9, -12, -6, 7, 16, 10, -9, -18, -11, 4, 15, 14, -2, -16, -20, -3,
14, 17, 6, -12, -24, -11, 10, 21, 14, -8, -22, -20, 3, 20, 17, -2, -21,
-24, -6, 19, 28, 10, -21, -36, -18, 19, 40, 22, -21, -42, -23, 16, 39,
26, -14, -40, -32, 9, 38, 29, -8, -39, -36, 2, 36, 38, -1, -38, -39, -6,
32, 42, 7, -34, -43, -14, 28, 46, 15, -30, -47, -22, 24, 50, 23, -26,
-51
%N A108618 A quaternion-generated sequence calculated using the rules
given in the comment box with initial seed x = .5'i + .5'j + .5'k + .5e
; version: "tes"    
%C A108618 Set y = x = .5'i + .5'j + .5'k + .5e  Define a(0) = 1 (this
is 2 times the coefficient of the unit e in x), then "loop" steps 1-5,
below. a(n) is given by 2 times the coefficient of e (the unit) in y
from step 4 inside of the the n-th loop.  

 
Step 1 (Loop 1): Calculate x*y
Result:  x*y =  .5'i + .5'j + .5'k - .5e

Step 2 (Loop 1): Add the fractional parts of the real coefficienct basis
vectors of x*y (i.e. 'i, 'j, 'k, e)
Result: .5 + .5 + .5 - .5 = 1 = s
   
Step 3 (Loop 1): Calculate x + x*y + se
Result  .5'i + .5'j + .5'k + .5e + (.5'i + .5'j + .5'k - .5e) + se
= 'i + 'j + 'k + e.

Step 4 (Loop 1): Set y equal to the result from Step 3.
Result: y = 'i + 'j + 'k + e ; thus a(1) = 2*1 = 2

Step 5 (Loop 1): Return to Step 1

 
Step 1 (Loop 2):
Result:  x*y = 'i + 'j + 'k - e

Step 2 (Loop 2):
Result: s = 0
 
Step 3 (Loop 2): 1.5'i + 1.5'j + 1.5'k -.5e

Step 4 (Loop 2): y = 1.5'i + 1.5'j + 1.5'k -.5e ; thus, a(2) = 2*(-.5) =
-1

 
**Loop 1**  + 'i + 'j + 'k + e
 
**Loop 2**  + 1.5'i + 1.5'j + 1.5'k - .5e
 
**Loop 3**  + 'i + 'j + 'k - e
 
**Loop 4**  + .5'i + .5'j + .5'k - 1.5e
 
**Loop 5**  - 3e
 
**Loop 6**  - 'i - 'j - 'k - 3e
 
**Loop 7**  - 1.5'i - 1.5'j - 1.5'k + .5e
 
**Loop 8**  + 2e
 
**Loop 9**  + 1.5'i + 1.5'j + 1.5'k + 1.5e
 
**Loop 10**  + 2'i + 2'j + 2'k
 
**Loop 11**  + 1.5'i + 1.5'j + 1.5'k - 2.5e
 
**Loop 12**  - 5e
%H A108618 C. Dement, <a
href="http://www.crowdog.de/home.html">The Floretions</a>.
%H A108618 C. Dement, <a
href="http://www.crowdog.de/QuaternionLoop.html">Graph of first 500
terms.</a>.
%Y A108618 Cf. A108619, A108620, A108621
%O A108618 0
%K A108618 ,sign,
%A A108618 Creighton Dement (crowdog at crowdog.de), Jun 12 2005
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RA 84.129.245.186
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