FullSimplify command does a good job for this.

Attachments:

Hello Mariusz,

I greatly appreciate your continual help on my question!

I looked at your two attached files (in your previous post). It is making great progress and the solution is nearly there! But, there is a small problem with the Maple/Mathematica program (probably, unexpected variable values somewhere in memory). It does not yield the correct values for b[k].

With regards to the question:

For integers k greater than or equal to 3, b[k] = b[k-1] + b[k-3] + 2*[.0794367(.460811)^((k-1)/2) + .255972(-.675131)^((k-1)/2) + .664591(3.21432)^((k-1)/2)], if k is odd; b[k] = b[k-1] + b[k-2], if k is even; b[0] = 1, b[1] = 4, b[2] = 5.

This piecewise defined recurrence relation is correct. I verified (using this piecewise defined recurrence relation, by hand calculations) the correct values: b[0]=1, b[1]=4, b[2]=5, b[3]=10, b[4]=15, b[5]=34, b[6]=49, b[7]=108, b[8]=157, b[9]=348, b[10]=505, etc.

If you have some time, can you look at your Maple/Mathematica program and see what is happening? We are so close to a "closed-form" solution to b[k] !!

I am greatly indebted to you for your valuable help.

Sincerely, Richard M. Low

Maybe so:

Solve for a[n]:

sol = First@ RSolve[{a[n] == 3*a[n - 1] + a[n - 2] - a[n - 3], a[0] == 1, a[1] == 2, a[2] == 7}, a[n], n];
Simplify[ToRadicals[(a[n] /. sol)], Assumptions -> n \[Element] Integers]

(*1/444 (2 (1 + (9 + I Sqrt[111])^(1/3)/3^(2/3) + 
  4/(27 + 3 I Sqrt[111])^(1/3))^
 n (74 + (20 74^(2/3))/(481 + 57 I Sqrt[111])^(
  1/3) + (74 (481 + 57 I Sqrt[111]))^(1/3)) + (1 + (
  I (I + Sqrt[3]) (9 + I Sqrt[111])^(1/3))/(
  2 3^(2/3)) + (-2 - 2 I Sqrt[3])/(27 + 3 I Sqrt[111])^(1/3))^
 n (148 + (20 I 74^(2/3) (I + Sqrt[3]))/(481 + 57 I Sqrt[111])^(
  1/3) + (-1 - I Sqrt[3]) (74 (481 + 57 I Sqrt[111]))^(
   1/3)) + (1 - ((1 + I Sqrt[3]) (9 + I Sqrt[111])^(1/3))/(
  2 3^(2/3)) + (-2 + 2 I Sqrt[3])/(27 + 3 I Sqrt[111])^(1/3))^
 n (148 - (20 74^(2/3) (1 + I Sqrt[3]))/(481 + 57 I Sqrt[111])^(
  1/3) + I (I + Sqrt[3]) (74 (481 + 57 I Sqrt[111]))^(1/3)))*)

Solve for w[n]:

sol2 = First@ RSolve[{w[2*n + 1] == (a[n] /. sol /. n -> n + 1), w[1] == 2, w[3] == 7, w[5] == 22}, w[n], n];
Simplify[ToRadicals[(w[n] /. sol2)], Assumptions -> n \[Element] Integers]

 (*1/222 (1 + (9 + I Sqrt[111])^(1/3)/3^(2/3) + 4/(27 + 3 I Sqrt[111])^(
     1/3))^((1 + n)/
   2) (74 + (20 74^(2/3))/(481 + 57 I Sqrt[111])^(
     1/3) + (74 (481 + 57 I Sqrt[111]))^(1/3)) + 
  1/444 (1 + (I (I + Sqrt[3]) (9 + I Sqrt[111])^(1/3))/(
     2 3^(2/3)) + (-2 - 2 I Sqrt[3])/(27 + 3 I Sqrt[111])^(1/3))^((
   1 + n)/2) (148 + (
     20 I 74^(2/3) (I + Sqrt[3]))/(481 + 57 I Sqrt[111])^(
     1/3) + (-1 - I Sqrt[3]) (74 (481 + 57 I Sqrt[111]))^(1/3)) + 
  1/444 (1 - ((1 + I Sqrt[3]) (9 + I Sqrt[111])^(1/3))/(
     2 3^(2/3)) + (-2 + 2 I Sqrt[3])/(27 + 3 I Sqrt[111])^(1/3))^((
   1 + n)/2) (148 - (
     20 74^(2/3) (1 + I Sqrt[3]))/(481 + 57 I Sqrt[111])^(1/3) + 
     I (I + Sqrt[3]) (74 (481 + 57 I Sqrt[111]))^(1/3))*)