Quite confused about two sequences

[creigh at o2online.de]

The following two mysteriously linked sequences, a(n) and b(n), should have 
meaning beyond the formula given below. Perhaps someone
could lend a comment about how best to proceed. I divided
them by 2 and 4 respectively because 
1. This was the way the program "spit them out" (not as arbitrary as one
would initially think).
2. a(2)/2 = -b(2)/4  =   3     =   3^1 
    a(8)/2 = -b(8)/4  = -81    =  -3^4
   a(14)/2 = -b(8)/4 = 2187  =   3^7
   are powers of 3 

a(n)/2 = (-0.5, -0.5, 3.0, -7.5, 13.5, -18.0, 13.5, 13.5, -81.0, 202.5, 
             -364.5, 486.0, -364.5, -364.5, 2187.0, -5467.5)

-> a(n) = (-1, -1, 6, -15, 27, -36, 27, 27, -162,  ... )

b(n)/4 = (-1.75, 2.75, -3.0, 0.75, 6.75, -22.5, 47.25, -74.25, 81.0, 
              -20.25, -182.25, 607.5, -1275.75, 2004.75, -2187.0, 546.75)

-> b(n) = (-7, 11, 12, 3, 27, -90,  189, -297, 324...)

A formula:
| a(n)/2 + b(n)/4 |  = (9/4)*A057681(n+1)
A057681 = (1,1,1,0,3,9,18,27,27,0,81,243,486,729,729,0,2187,6561,13122, )
http://www.research.att.com/projects/OEIS?Anum=A057681

An example of formula:
| a(7)/2 + b(7)/4 | = | 13.5 - 74.25 | = 60.75 = (9/4)*A057681(8)

Conjectures: 
1. a(n)/2 +  b(n)/4 = 0  -> exists m: | a(n)/2 | = | b(n)/4 | = 3^m
2. forall n exists m :  ( | b(n)/2   -  a(n)/2 |  = 3^m or | b(n)/2   +  
a(n)/2 | = 3^m ) if and  only if | a(n+1)/2 | != | b(n+1)/4 |
3. |a(n)| = |a(m)| -> a(n) = a(m)
 
A reminder- the formulas for a(n) and b(n) may well be connected to 
the Fibonacci (or related) numbers due to the similarties in appearance 
among the following three sequence formula. This, in turn, would 
effectively connect "The Smith College diploma problem",
i.e. A057681, to the Fibonacci numbers via the formula given above.
(it would be interesting to know what the Smith problem was-
however, the link given at OEIS appears to be inactive and I haven't 
had time to search beyond that).

A057681(n)  = ves( (E'x)^n )/2 
E' = ( - 'i - i' + 'ii' + 'jj' + 'kk' + 'jk' + kj' + 1 )/4
x = 'i + i' + 'ji' + 'ki' + 1 (signs are preserved) 

A000032(n) = L(n) = ves( (Ex)^n ) (disregarding signs) 
E = ( 'i + i' + 'ii' + 'jj' + 'kk' + 'jk' + kj' + 1 )/4
x = 'i + i' + 'ji' + 'ki' + 1 

A000045(n) = Fib(n) = ves( (E''x)^n ) 
E'' = ( - 'i - i' - 'ii' + 'jj' + 'kk' + 'jk' + kj' - 1 )/4
x = 'i + i' + 'ji' + 'ki' + 1 

"Superseeker" did not have an answer for a(n):
*******************************
Report on [ 1,1,6,15,27,36,27,27,162]: 

Even though there are a large number of sequences in the table, at least one 
of yours is not there! Please send it to me using 
the submission form on the sequence web page 
http://www.research.att.com/~njas/sequences/Submit.html 
and I will (probably) add it! Include a brief description. Thanks! 
[snip]
Guesss suggests that the generating function F(x) 
may satisfy the following algebraic or differential equation: 

x^5-3/7*x^4+2/7*x^3+1/3*x^2-1/9*x+1/18+(5/7*x^4-2/7*x^3-1/6*x^2+1/6*x-1/18)*F(x) = 
0 

If this is correct the next 6 numbers in the sequence are: 

[729, 13365/7, 21384/7, 12393/7, -210681/49, -65610/7] 

("729" does turn up as 2*(364.5) = a(10). However, the program
apparently thinks 729 = a(9) ).  

Thanks for any suggestions, 
Creighton  http://www.crowdog.de