a(n)=Number of Unique Matrix Products in (A+B+C)^n When {A,B}=0

[Paul D. Hanna]

Christian (and Seqfans), 
      Very interesting statements! 
Your proposed g.f. I believe captures the underlying principle: 
> 1/(1-a1*x+a2*x^2-a3*x^3+...) where
> a1 is number of letters
> a2 is number of commuting pairs
> a3 is number of commuting triples
I did notice that the g.f.s were of your simple form, but 
failed to see the significance of the coefficients.  
 
To throw a wrench in the works, one could also introduce  
letters that ANTI-COMMUTE.   Common in physics.  
The number of distinct sets of n-letters then would be different  
since some parts would cancel.  
Example:  if A,B anti-commute, {A,B}=AB+BA=0,  
then (A+B)^n has 2^[(n+1)/2] distinct parts (g.f. (1+2x)/(1-2x^2)). 
 
Here are 3-letter cases begging enumeration (and a g.f.): 
  
(1) What is the number of distinct parts in (A+B+C)^n  
when A,B,C anti-commute with each other?  
{A,B}={B,C}={C,A}=0 (mutually orthogonal case). 
  
(2) What is the number of distinct parts in (A+B+C)^n 
when A,B anti-commute and A,B commute with C? 
{A,B}=0, [A,C]=[B,C]=0.
 
(3) What is the number of distinct parts in (A+B+C)^n 
when A,B anti-commute and A,B do not commute with C? 
{A,B}=0, [A,C] !=0, [B,C] !=0.
   
(4) What is the number of distinct parts in (A+B+C)^n 
when A,B anti-commute and A,C commute but B,C do not commute? 
{A,B]=0, [A,C]=0, [B,C] !=0.
   
The interaction between commutative, non-commutative, and
anti-commutative 
letters would be nice to have encapsulated in one simple g.f. like one
you proposed. 
But we can save that one for a very rainy day. 
 
Thanks for the nice analysis. 
       Paul