Ternary Gray code

Enter "Gray code conversion rules" into Google to bring up the general rules; following rules in A147995 in OEIS: http://www.tinyurl.com/4zq4q
then, following the reflection principle for Gray code k, we write out (say k=3) 0,1,2; 2,0,1; 1,2,0; ...in the right column; i.e. we write each term in the series once, then for the next subset of 3, we repeat the last term recorded, or "reflect" it.
...0
...1
...2;
now reflect, starting the series again (2, then 0, 1,...):
...2,
...0,
...1;
now for the next subset we reflect the "1" getting
...1
...2 (continuing with 0,1,2, etc.)
Now for the next column going to the left, we mark down each term 3 times; and the next column after that (column 3), 9 times, so that each term in the series 0,1,2 (and generally, 0,1,2,3,...for any k) is marked down k^0, k^1, k^2,....times along with the reflection principle. Thus, for Ternary Gray code we obtain:
000
001
002
012
010
011
021
022
020
120
121
122
102
101
100
110
111
112
...

Gray code conversion rules

at http://www.tinyurl.com/4zq4q
then access sequence A147995
The reflection principle is as follows, by way of example, ternary Gray code :
First column, going right to left, within each set of 3 terms starting with 0, write 0,1,2, then the next set repeats the 2 (reflection) then continues with the series 0,1,2. Each set reflects, then continues in the series 0->1->2, so we obtain
0,1,2; 2,0,1; 1,2,0; 0,1,2;...; (but these are column 1 terms.). Column 2 terms each term is marked down 3 times: 0,0,0; then 1,1,1; 2,2,2; then reflect: 2,2,2;...
For column 3, each term is written down 9 times along with the reflection principle as before. Generally for Gray code K, each term is recorded K^0, K^1, K^2...times by columns. Putting the rules together for k=3, Ternary, we
obtain (for decimal 0, 1, 2, 3,....):
000
001
002
012
010
011
021
022
020
120
121
122
102
101
100
110
111
112
...
and, we note that the column change from n-th to (n+1)-th term =
A051064 in OEIS: (1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, ...) representing
the row (labeled 1,2,3,...) in the following multiplication table:
(heading terms not multiples of 3, * left column = powers of 3):
1,...2,...4,....5,....7,...8,...10...
3,..6..,12,..20, 21,.24,..30,..
9,..;
Then record the row that n=1,2,3,...occurs in, getting A051064 as before:
(1, 1, 2, 1, 1, 2, 1, 1, 3,...) = the ruler sequence for k=3.

Genomic matrices

Generalized Genomic Matrices, Silver Means, and Pythagorean Triples"> > is now published on the web;> http://www.scipress.org/journals/forma/frame/24.html>

partitions and the formal power series

But to review, you have given p(x) = polcoeff partition series = A(x)/A(x^2), = B(x)/ B(x^3)....; such that A(x), B(x),...etc for any k = Euler transform of the generalized ruler sequences ("p-adic valuations of kn"). Below we have an important algebraic operation and identity.It states that given A(x), B(x), etc,....for any k = 2, 3,...; A(x) = p(x) * p(x^2) * p(x^4) * p(x^8) * p(x^16) *....; andB(x) = p(x) * p(x^3) * p(x^9) * p(x^27) * p(x^81) * ...etc. Gary
Date: Wed, 17 Mar 2010 17:16:59 -0700From: qntmpkt@yahoo.comSubject: Fw: [seqfan] Re: need algebra proofTo: qntmpkta@gmail.com; qntmpkta@hotmail.com
Conjecture:> Given polcoeff S(x), by way of example given A000041, the partition sequence = (1, 1, 2, 3, 5, 7, 11, 15, 22,...); then> S(x) = (1 + x + 2x^2 + 3x^3 + ...).>> Then S(x) = some A(x) such that S(x) = A(x) / A(x^2).
If you replace the somehow obscure " S(x) = some A(x) " (?)by "there exists A(x)", then this is indeed true,given the hypothesis that S(0) = 1.(Else such an A cannnot exist, because A(0) / A(0) = 1 (or ill definedif A(0)=0).)
> Our conjecture states that A(x) = S(x) * S(x^2) * S(x^4) * S(x^8) * S(x^16) * ...).
Obviously, A(x) is not uniquely defined, so you should say "onesolution is given by ...".The choice of this solution is equivalent to the assumption (or choice) A(0)=1.You could multiply this A by any nonzero constant and you would againget the same A(x) / A(x^2).
The infinite product converges in the topology of formal power series(which means that the coefficients up to a given orderare only affected by a finite number of terms, cf.http://en.wikipedia.org/wiki/Formal_power_series )so it is perfectly well defined.To exclude any ambiguity, you can specifythat the expression ... / A(x^2)means ... * ( A(x^2) )^(-1) ,where ( 1 - T(x) )^(-1) = 1 + T(x) + T(x)^2 + ...(where T(x) is any power series with T(0)=0).
So you can check that the initial formula is true(which means : coefficients of any given power are the sameon the l.h.s. and on the r.h.s.)by writing out the relevant factors of the infinite productsand cancelling those which correspond.
> Similarly, for S(x) = B(x) / B(x^3), our conjecture states that:>> B(x) = S(x) * S(x^3) * S(x^9) * S(x^27) * S(x^81) * ...>> Generally, for any k, in S(x) = C(x) / C(x^k).>
Yes, this is the same, the only necessary condition is again that S(0)= 1, and assuming the choice B(0) = C(0) = 1.
Maximilian
PS: I think Richard Mathar already pointed out roughly the same thing

Partitions and the formal power series

https://mail.google.com/mail/?shva=1#inbox/1276eab0f83c07c3

Hogan's quantum noise

at http://astro.fnal.gov/people/Hogan

Gary W. Adamson articles

https://mail.google.com/mail/?shva=1#inbox