Why isn't Mathematica giving me the desired output for this RSolve equation?

[ehrenfest]

Homework Statement

When I type

RSolve[{a[n] == a[n - 1] + a[n - 2], a[1] == 1 && a[2] == 1}, a[n], n]

into mathematica 6 why does it give me

{{a[n] -> Fibonacci[n] - C[2] Fibonacci[n] + C[2] LucasL[n]}}

and not just

{{a[n] -> Fibonacci[n] }}

?

Homework Equations

The Attempt at a Solution

 

[NateTG]

Could be an off-by-one. What does:

RSolve[{a[n] == a[n - 1] + a[n - 2], a[1] == 1 && a[0] == 1}, a[n], n]

give?

 

[ehrenfest]

same thing.

{{a[n] -> Fibonacci[n] - C[2] Fibonacci[n] + C[2] LucasL[n]}}

I think its a bug.

 

[NateTG]

Hmm, is the syntax you're using correct? (I'm not a mathematica user.)

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

or

RSolve[{a[n] == a[n - 1] + a[n - 2] &&  a[1] == 1 && a[0] == 1}, a[n], n]

seem more consistent.

 

[R.Harmon]

I tried it and it just gives {{a[n] -> Fibonacci[n] }} for me. Not sure why, my mathematica knowledge isn't that great, but like you said could just be a bug?

 

[EnumaElish]

In Math'ca 6.0.1.0,

RSolve[{a[n] == a[n - 1] + a[n - 2], a[1] == 1 && a[2] == 1}, a[n], n]

gives

{{a[n] -> Fibonacci[n]}}

whereas

RSolve[{a[n] == a[n - 1] + a[n - 2], a[1] == 1, a[0] == 1}, a[n], n]
and
RSolve[{a[n] == a[n - 1] + a[n - 2] && a[1] == 1 && a[0] == 1}, a[n], n]

both give

{{a[n] -> 1/2 (Fibonacci[n] + LucasL[n])}}

 

[ehrenfest]

Yes. You're right. I just did a Clear["Global`*"] and began getting the same answer as you. I am not really sure how why the Clear["Global`*"] would change the evaluation of this though...it seems like all variables should be local.