Lucas Strings

[Creighton Dement]

Dear Seqfans, 

I've submitted the previously discussed sequence -which seems like
should be "of importance".  Much time was spent attempting to formulate
a concise yet readable explanation- which I tried to do without defining
"a" and "b", below, as floretions.  Regardless, I can hardly claim that
my explanation is the best. In fact, I can't even claim that it's
particularly good (in other words,  suggestions welcome! ) .
  
%S A000001 0, 1, 1, 3, 3, 8, 8, 17, 23, 41, 55, 102, 144, 247, 387, 631,
987, 1636
%N A000001 The number of a's in the partition of tes[(a+b)^n], shown
below (where n > 0). The number of b's minus the number of a's returns
A034748: Dirichlet convolution of Fibonacci numbers with phi(n).
%C A000001 Note: The paper "Floretion-generated Integer Sequences" (in
progress) gives a more detailed definition of (a(n)) (see chapter:
Necklaces).

Let R({a,b}) be a ring generated by two elements a and b such that a^2 =
0 ; bx != 0 and xb != 0 for all non-zero x in R({a,b}). Let  tes:
R({a,b}) -> reals be a function with the following properties: I:
tes[xy] = tes[yx] for all x, y in R({a,b}). II: tes[x+y] = tes[x] +
tes[y]. III: tes[a] = tes[a^2] = 0. IV: tes[b] != 0. V: if x in 
R({a,b}) and bx != 0, then tes[bx] != 0. Note: To define (a(n)), it is
only important that the range of tes is a ring (not necessarily the
reals). Set Lucas = A000204.

Then (a+b)^1 = a+b (number of non-zero terms = Fib(3)), tes[a+b] =
tes[a] + tes[b] = tes[b] (number of non-zero terms = Lucas(1));
(a+b)^2 = (a+b)(a+b) = a^2 + ab + ba + b^2 = ab + ba + b^2 (number of
non-zero terms = Fib(4)), tes[(a+b)^2] = tes[ab + ba + b^2] = tes[ab] +
tes[ba] + tes[b^2] = 2*tes[ab] + tes[b^2] (number of non-zero terms = 3
= Lucas(2));
(a+b)^3 = (ab + ba + b^2)(a+b) = aba + ba^2 + b^2a + ab^2 + bab + b^3 =
aba + b^2a + ab^2 + bab + b^3 (number of non-zero terms = Fib(5));
tes[(a+b)^3] = tes[aba] + tes[b^2a] + tes[ab^2] + tes[bab] + tes[b^3] =
tes[b^2a] + tes[ab^2] + tes[bab] + tes[b^3] = 3*tes[b^2a] + tes[b^3]
(number of non-zero terms = 3 = Lucas(2)). Continuing, it can be shown
that the number of non-zero terms is always given by Lucas(n). To
calculate a(n), partition tes[(a+b)^n] as above. Next, disregard the
leading coefficients and count the total number of a's.  Example:  The
total number of b's in the above partition of tes[(a+b)^3], ignoring
leading coefficients, is 5. The total number of a's is 1.  It follows
that a(3) = 1. Furthermore, A034748(3) = 5 - a(3) = 4.
%D A000001 C. Dement, Floretion-generated Integer Sequences (work in
progress).
%H A000001 C. Dement, <a
href="http://www.crowdog.de/13829.html">The Floretions</a>.
%F A000001 Conjecture: a(p) = Fib(p-1) for all primes, where Fib =
A000045 (Creighton Dement and Antti Karttunen).
%e A000001 a(1) = 0 since tes[a+b] = tes[b] and the number of a's in
tes[b] is 0.
%Y A000001 Cf. A034748, A006206, A000358, A000045, A000204.
%O A000001 1,4
%K A000001 ,more,nice,nonn,
%A A000001 Creighton Dement (crowdog at crowdog.de), Jan 05 2006







-It's a shame when the girl of your dreams would still rather be with
someone else when you're actually in a dream.