Changes made to a floretion to induce an array of binomial transforms
creigh at o2online.de
creigh at o2online.de
Sun Nov 7 14:33:33 CET 2004
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).
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