Perhaps, a strange property of Pell and friends.

[creigh at o2online.de]

Greetings! 

I begin by defining the following sequences, whose bisections are all listed 
in OEIS (each of these bisections seems to satisfy the same recurrence relation 
a(n) = -6a(n-1) - a(n-2) ; the denominators of the generating functions 
being identical). Surprisingly, the denominators of 
each of the sequences c, d, e, f, g and h are identical (a conjecture, 
based on empirical evidence, as to why this is so is also given). 

(c(n)) = (1, -3, -7, 17, 41, -99, -239, 577, 1393, -3363, -8119, 19601, 47321, 
)
( This is the sequence "Numerators of continued fraction convergents to sqrt(2)", 
apart from signs. ) Taking signs into account, we have g.f : -(x^3+x^2+3*x-1)
/(x^4+6*x^2+1)
Disregarding signs, the g.f. given is: g.f.: (1-x)/(1-2*x-x^2).

c(2n) = ((-1)^n)*A001653(n),
( NSW numbers: a(n) = 6*a(n-1) - a(n-2); also a(n)^2 - 2*b(n)^2 = -1 with  
b(n)=A001653(n) )
 c(2n+1) = ((-1)^(n+1))*A001541(n+1)
(A001541, Chebyshev polynomials of the first kind evaluated at 3)

(d(n)) = (2, 4, -10, -24, 58, 140, -338, -816, 1970, 4756, -11482, -27720, 66922, 
) 
 (twice Pell numbers, apart from signs)
d(2n) = ((-1)^n)*A075870(n),
(A075870, 2*n^2 - 4 is a square )
d(2n+1) = ((-1)^n)*A005319(n+1)
(A005319,  http://www.research.att.com/projects/OEIS?Anum=A005319 ; a(n) 
= 6a(n-1) - a(n-2), see comment on this sequence ) 

  
(e(n)) =  (1, -1, -5, 5, 29, -29, -169, 169, 985, -985, -5741, 5741, 33461, -33461  
) 
(accounting for signs) g.f. - (x^2+1)(x-1)/(x^4+6*x^2+1)
e(2n) =  d(2n)/2,  e(2n+1) = - d(2n)/2

(f(n)) = (2, 2, -12, -12, 70, 70, -408, -408, 2378, 2378, -13860, -13860, ) 
f(2n) =  d(2n+1)/2,  f(2n+1) =  d(2n+1)/2

(g(n)) = (0, -3, 0, 17, 0, -99, 0, 577, 0, -3363, 0, 19601, 0, -114243, 0, 
665857)
g(2n) = 0, g(2n+1) = c(2n+1)

(h(n)) = (3, 4, -17, -24, 99, 140, -577, -816, 3363, 4756, -19601, -27720,  )
h(2n) = - c(2n+1), h(2n+1) = d(2n+1)

In my opinion, there are (at least) four interesting points about this:

1. The denominators of the generating functions of the sequences c through h 
are apparently the same: (x^4+6*x^2+1). On an empirical basis, this is to 
be expected as these are "static symmetries" of type "em" - however, a general 
proof that the den. of the generating functions attributed to static symmetries 
are identical lacks entirely and may be shown to be either trivial or extremely 
complicated to prove in the end. The next point is similar:

2. The sign make-ups of c, e and g as well as of d, f, and h are identical. c, 
e, and g belong to the symmetries "em[I], em[J], fam", respectively. d, 
f and h belong to the symmetries "em[I*], em[J*], fam* " (note: "fam"  also 
belongs to the "em" class of symmetries).

3. c(n) + d(n) = e(n) + f(n) = g(n) + h(n)  (A conjecture. Actually, when dealing 
with the floretions, it is easy to forget that - at this stage - almost 
everything one does is part of some greater conjecture. This includes taking 
the floretion 'i + j' + k', incidently the generating 
floretion for the relations given on this page, and stating that one of 
its resulting sequences x has generating function y... because, in actuality, 
the sequence stops at some point and one is always forced to guess(s) that 
its g.f. is this or that.)

4. c(n) + d(n) = b(n), b(2n) = c(2n+1), b(2n+1) = c(2n). In other words, the 
sequences (c(n) + d(n)) = (e(n) + f(n)) = (g(n) + h(n)) all represent the 
sequence c with even and odd indexed terms reversed! 

Unless new sequences e, f, g, h and b are defined, this would make for quite 
a long "Comment" (for example, on the sequence "Numerators of continued 
fraction convergents to sqrt(2)"). Any tips?

Sincerly, 
Creighton