Dear @Richard Low, please spend some time learning how to properly post math notation and code on Wolfram Community:

http://community.wolfram.com/groups/-/m/t/270507

Hello Mariusz,

I'm very glad to see your attached file MMASolution-Ver4.nb, which gives a "closed-form" solution to b[k] for ODD numbers! Thank you!!

1. In this particular "closed-form" solution (for ODD numbers), is it possible to replace all of the complicated (exact) radical expressions with simpler decimal approximations? This would yield a very compact formula which could be typeset into the research paper.

2. Do you have any ideas for a "compact" (with simpler decimal approximations, instead of exact radical expressions) formula for b[k], for EVEN numbers?

Again, thank you for all of your help given and time spent.

Sincerely, Richard M. Low

This is extremely confusing. For one, the notation is not correct Mathematica so nothing can be cut-and-pasted directly. Moreover, the thread suffers from "question creep", whereby the specifics keep changing to the point where it is difficult to figure out any more what is the question under consideration. Also there seems to be no effort, or at least none shown, to adapt responses into actual RSolve code (this does not add to the confusion, but is nonetheless discouraging for what I think are obvious reasons).

Here is something that probably does what is needed, albeit in I am sure the wrong notation (but see remark on "question creep"). We get away from the odd/even issue by separating into two functions; this idea also seems to have been in some of the prior posts (again, see remark on "question creep").

a[n_] := .0794367 (.460811)^n + .255972 (-.675131)^
     n + .664591 (3.21432)^n;

rv = 
   RSolveValue[{b[n] == c[n] + c[n - 1] + 2*a[n], 
     c[n] == b[n - 1] + c[n - 2], c[0] == 1, c[1] == 5}, {b[n], c[n]}, n]

  (* Out[204]= {2.9999999999999982*2.^n + 
    Piecewise[{{(0.32057243726157714*(-675131.)^n + 
          0.055654497058256*460811.^n - 
                  1.7999987988497759*E^(14.508657738524217*n) + 
          2.423771564529943*E^(14.983126384729113*n))/
               E^(13.815510557964275*n), 
       n > 1.}}, (0.5119440000000001*(-675131.)^n + 
        0.15887340000000003*460811.^n + 
              1.329182*E^(14.983126384729113*n))/
      E^(13.815510557964275*n)] - 
    2.9999999999999982*2.^n*UnitStep[-1.*n] + 
       C[1]*
     UnitStep[-1.*
       n], (-1.0000005443826219*(-0.3977012401567452*(-675131.)^n*
         E^(0.7747672983575988*n) + 
             0.03256025590597696*E^(13.815510557964275*n) + 
        1.*(-1.)^n*E^(14.590277856321874*n) - 
             1.9999989112353476*2.^n*E^(14.590277856321874*n) + 
        1.1999985459748292*E^(15.283425036881816*n) - 
             0.8348581061063877*E^(15.757893683086712*n)))/
    E^(14.590277856321874*n)} *)

Check the result numerically:

Table[rv, {n, 0, 6}]

(* Out[206]= {{1.9999994 + C[1], 1.}, {9.9999976319, 5.}, {29.9999920347,
   10.9999976319}, {89.9999745305, 34.9999920347}, {277.999927528, 
  100.999972162}, {869.999809803, 312.999919563}, {2749.99955564, 
  970.999781965}} *)

With RSolve you solve for a(n), and then w(k)=a((k-1)/2). Finally you check whether the initial conditions for w are met. To get you started:

RSolve[{a[n] == 3*a[n - 1] + a[n - 2] - a[n - 3], a[0] == 1, 
   a[1] == 2, a[2] == 7}, a, n]

You really should show what you tried. It's not the sort of thing I'd want to guess at.