[seqfan] Re: detective work related to Creighton Dement's \\\"Floretions\\\"

Creighton Kenneth Dement creighton.k.dement at mail.uni-oldenburg.de
Fri Nov 20 10:31:49 CET 2009

>> here needs to be:
>
> 1) A permanent web site explaining all the notation and computations.
> (There ought to be someplace to put such documentation in the Wiki
> OEIS.)
>
> 2) A link from every floretion-generated sequence to this web site.
>
>
> Me:  I agree!  Right now it is trivial for me to
> add a link to a list of sequences.  (It might not
> be so trivial once we go wiki.)  If someone
> sends me a list of A-numbers and the link,
> I have a program that will do the rest of the work.
>
> Neil

I totally agree.

Here are the very basics which should be any such page.

Let X = A'i  + B'j + C'k + Di' + Ej' + Fk' + G'ii' + H'jj' + I'kk' +
J'ij' + K'ik' + L'ji' + M'jk' + N'ki' + O'kj' + P'ee'

and Y = a'i  + b'j + c'k + di' + ej' + fk' + g'ii' + h'jj' + i'kk' +
j'ij' + k'ik' + l'ji' + m'jk' + n'ki' + o'kj' + p'ee'

Then

ibase(X) = A
jbase(X) = B
kbase(X) = C
basei(X) = D
basej(X) = E
basek(X) = F
ibasei(X) = G
jbasej(X) = H
kbasek(X) = I
ibasej(X) = J
ibasek(X) = K
jbasei(X) = L
jbasek(X) = M
kbasei(X) = N
kbasej(X) = O
tes(X) = P

jesleft(X) = A+B+C
jesright(X) = D+E+F
jes(X) = A+B+C+D+E+F
les(X) = G+H+I+J+K+L+M+N+O

We have the identities:
ves(X) = jes(X) + les(X) + tes(X)
jes(X) = jesright(X) + jesleft(X)

vesseq(X) = (ves(X), ves(X^2), ves(X^3), ...)
Note: multiplication is defined below.

Similarly,
jesseq(X) = (jes(X), jes(X^2), jes(X^3), ...)
lesseq(X) = (les(X), les(X^2), les(X^3), ...)
tesseq(X) = (tes(X), tes(X^2), tes(X^3), ...)

ibasepos(X) = A, if A positive
= 0 otherwise [ alternatively, ibasepos(X) = (A + |A|)/2 ]

ibaseneg(X) = A, if A negative
= 0 otherwise [ alternatively, ibaseneg(X) = (A - |A|)/2 ]

Thus, ibasepos(X) + ibaseneg(X) = ibase(X)

Following the same pattern as above, we define:
jesleftpos(X) = ibasepos(X) + jbasepos(X) + kbasepos(X)
jesrightpos(X) = baseipos(X) + basejpos(X) + basekpos(X)

vespos(X) = ibasepos(X) + ... + kbasejpos(X) + tespos(X)

X*Y =

(Pa - Dg - Ho + Im - Fk - Ej + Oh - Cb - Gd + Nl - Je - Kf - Mi + Ap + Bc
- Ln)'i

+

(-Mf + Bp - Ac - Ng - Ld - Ik + Jo + Gn - Dl - He + Pb - Oj + Ki - Eh - Fm
+ Ca)'j

+

(Ab - Jh - Eo - Dn - Km + Cp - Nd + Lg + Mk - Gl - Fi - If + Hj - Ba + Pc
- Oe)'k

+

(Kj - Fe + Io - Hm - Jk + Mh - Ag - Lb - Ga - Cn + Dp - Nc - Oi + Ef - Bl
+ Pd)i'

+

(Fd - Ml - Bh - Co - Oc + Ni - Hb - Kg - Aj + Ep - In + Lm + Gk + Pe - Df
-Ja)j'

+

(Pf - Ed - Ci + Fp - Lh + Jg - No + On - Ic - Mb - Bm - Ka + De - Ak - Gj
+ Hl)k'

+

(Bn + Da + Ad + Pg - Cl + Jf + Gp + Ek + Hi - Om + Lc - Mo + Ih - Nb - Fj
- Ke)'ii'

+

(-Nk - Dm + Ph + Eb - Lf - Ao + Cj + Oa - Kn - Jc + Gi + Fl + Be + Hp + Md
+ Ig)'jj'

+

(-En + Ip - Bk + Pi + Am + Do + Cf + Fc - Ma + Ne - Jl - Lj - Od + Gh + Kb
+ Hg)'kk'

+

(-Ch - Li - Dk - Il + Ea + Mn + Kd + Pj - Ob + Ae + Hc + Nm + Jp + Bo - Gf
+ Fg)'ij'

+

(Ol - Hn + Lo - Eg - Nh - Ib + Af - Jd + Bi - Cm + Kp + Dj + Pk + Ge + Fa
+ Mc)'ik'

+

(-Gc + Na - An + Em - Ji + Db + Pl - Fh - Ij + Ok + Bd + Cg + Ko + Hf + Lp
- Me)'ji'

+

(Le + Bf + Ck + Mp + Nj - El + Jn - Go + Ia + Dh + Fb + Pm - Hd - Ai - Og
-Kc)'jk'

+

(-Bg + Mj + Of + Cd + Gb - La + Np + Al + Ei - Fo + Dc - Ie - Kh + Pn + Jm
- Hk)'ki'

+

(Ah - Mg + Kl - Bj + Ec - Ha - Gm - Nf - Di + Lk + Fn + Ce + Op + Id + Jb
+ Po)'kj'

+

(-aA - bB - cC - dD - eE - fF + gG + hH + iI + jJ + kK + lL + mM + nN + oO
+ pP)'ee'

******************

It can be shown that adding the unit vector 'ee' to a floretion X induces
a set of sequences which represents the binomial transforms of the
original set generated by X. In particular, if

(a(n)) = (1, tes(X), tes(X^2), ...), then
(b(n)) = (1, tes(X+'ee'), tes((X+'ee')^2), ...) is the binomial transform
of X.

Seqfan note:
It doesn't get much simpler than the definitions above, but we can see how
quickly difficult questions arise. How can you prove, for example, that
vesposseq(X) is always given by a linear recurrence relation of a certain
order. In almost every case, I believe it is order 16 or below.

I should stress that floretions have been an "after-hours" hobby of mine
for the last several years. Until I can find some other way to pay the
rent, that will remain so for the foreseeable future. This means that
there is currently no possibility for me to write an entire book of
documentation (which by now I probably could).

Mr Munafo wrote:
*********
After a while I started down-rating sequences like
A100545 and A113166 that say they were floretion-generated but provide no
further clues about the floretion definition or algorithm.
*********

A100545 was explained in my last post. A113166 is another one of my
favorite sequences. I have already explained how this sequence was
generated. See
http://www.research.att.com/~njas/sequences/a113166.txt
You won't find a direct reference to floretions there, but floretion
multiplication was used (it's just a matter of me looking back up which
floretions were involved and I will do that). Refer also to the text
directly following proposition 2.5 in the draft paper.

As for ambiguities in my (older) sequences, I see two cases:

1. Minor ambiguity: Here (see, for ex. A107849 or A113166) the sequence is
properly defined without using floretions but some reference or comment is
made that either the idea for the sequence came from floretions or code is
present which cannot be directly executed.

Quick solution (though very unfortunate and I apologize if this is
time-consuming): Remove the code and/or references to floretions.

2. "Major ambiguity" (rare, such as the themesong sequence or the very
first batch of sequences I ever submitted)

Quick solution: delete the sequence (it was probably only a representative
sequence taken from a batch of hundreds of others- and I can still
reconstruct the process, pick another representative, and submit all
documentation at a later time)

Sincerely,
Creighton