Changes made to a floretion to induce an array of binomial transforms

[creigh at o2online.de]

Hi!

Let 

x = (x1) 'i + (x2) 'j +  (x3) 'k +  (x4) i' + (x5) j' +  (x6) k' +  (x7) 'ii' 
+  (x8) 'jj' +  (x9) 'kk' +  (x10) 'ij' +  (x11) 'ik' +  (x12) 'ji' +  (x13) 
'jk' +  (x14) 'ki' +  (x15) 'kj' +  (x16) e

be any floretion which induces an array of integer sequences by summing up 
over basis vectors in accordance with some rule (i.e. "symmetry") [x1 ... 
x16 are real numbers]. 

For any natural number n, indications (see below) point to the property that 
the array of sequences induced by x + (sign(x16))*ne are also integers and 
are in fact quite likely to lead to an array of "nth binomial transforms" 
of the first set of sequences. 

I initially proposed this a few months ago (but am just now getting back around 
to it).  I do not give the whole array of sequences, below (for
the particular floretions I chose as examples) but choose to only look 
at ves, or at jes or pos). I presume a proof of the proposition would be 
somewhat lengthy and probably based on induction.

Sincerely, 
Creighton 

n = 0: 
0.25'i + 0.5'j - 0.75'k - 0.25i' - 0.25k' + 0.5'ii' + 0.25'ij' + 0.75'ji' 
- 0.25'jk' + 0.5'ki' + 0.25'kj' + 0.5e

vesseq: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,

n = 1:  
 + 0.25'i + 0.5'j - 0.75'k - 0.25i' - 0.25k' + 0.5'ii' + 0.25'ij' + 0.75'ji' 
- 0.25'jk' + 0.5'ki' + 0.25'kj' + 1.5e 

 vesseq: 3, 8, 20, 48, 112, 256, 576, 1280
http://www.research.att.com/projects/OEIS?Anum=A001792
Binomial transform of natural numbers [1,2,3,4,...].


******************* (new floretion)
n = 0:

0.5'j + 0.5j' - 0.5k' - 1'ii' + 1'kk' - 0.5'ji' + 0e 
posseq: 2, 4, 6, 12, 18, 36, 54, 108, 162
http://www.research.att.com/projects/OEIS?Anum=A068911
Number of n step walks (each step +/-1 starting from 0) which are never
              more than 2 or less than -2.

n = 1:
 0.5'j + 0.5j' - 0.5k' - 1'ii' + 1'kk' - 0.5'ji' + 1e 
posseq: 3, 9, 25, 69, 189, 517, 1413, 3861,
http://www.research.att.com/projects/OEIS?Anum=A077846
Binomial transform of A068911

***************
n = 0:
 - 0.25'i + 0.5'k - 0.25i' - 0.5j' + 0.5k' - 0.75'ii' + 0.75'jj' - 0.25'kk' 
+ 0.25'jk' - 0.5'ki' + 0.25'kj' + 0.25e 
4lesseq: -1, -1, -2, -3, -5, -8, -13, -21, -34, -55, -89, -144,  // Fib

n = 1:
 - 0.25'i + 0.5'k - 0.25i' - 0.5j' + 0.5k' - 0.75'ii' + 0.75'jj' - 0.25'kk' 
+ 0.25'jk' - 0.5'ki' + 0.25'kj' + 1.25e 
4lesseq: -1, -3, -8, -21, -55, -144, -377, 
Binomial transform of A000045

n = 2: 
 - 0.25'i + 0.5'k - 0.25i' - 0.5j' + 0.5k' - 0.75'ii' + 0.75'jj' - 0.25'kk' 
+ 0.25'jk' - 0.5'ki' + 0.25'kj' + 

2.25e 
4lesseq: -1, -5, -20, -75, -275, -1000, -3625
2nd binomial transform of Fib(n)

n = 3
 - 0.25'i + 0.5'k - 0.25i' - 0.5j' + 0.5k' - 0.75'ii' + 0.75'jj' - 0.25'kk' 
+ 0.25'jk' - 0.5'ki' + 0.25'kj' + 

3.25e 
 4lesseq: -1, -7, -38, -189, -905, -4256,
Associated to the knot 8_12 by the modified Chebyshev transform
              A(x)-> (1/(1+x^2)^2)A(x/(1+x^2)). See A099454 and A099455.
Links:     The Rolfsen Knot Table
Formula:   a(n)=sum{k=0..floor(n/2), binomial(n-k,k)(-11)^k*7^(n-2k)}.

n = 4
 - 0.25'i + 0.5'k - 0.25i' - 0.5j' + 0.5k' - 0.75'ii' + 0.75'jj' - 0.25'kk' 
+ 0.25'jk' - 0.5'ki' + 0.25'kj' + 4.25e 

4lesseq: -1, -9, -62, -387, -2305, -13392, -76733

http://www.research.att.com/projects/OEIS?Anum=A081574
Fourth binomial transform of Fibonacci numbers F(n).