# Recurrence relation word problem

Suppose that you have a large supply of red, white, green and blue poker chips. you want to make a vertical stack of n chips in such a way that the stack does not contain any consecutive blue chips.

Find a recurrence relation for $a_n$ where $a_n$ denotes the number of ways you can make such a stack of n poker chips.

Blue chip first then i have 3 choices for next chip of the n-2 remaining symbols i write ($a_{n-2}$)?

for the second case where blue is not the first chip simply let ( $a_{n-1}$ ) be the Operation? symbol? function? i dont know on the remaining n-1 symbols.

$a_n = 3( a_{n-1} )+ 3( a_{n-2} )$

Does this result make any sense?

• The result aside, the way in which you wrote the result and your thoughts makes little to no sense. In any case, note that a stack with no consecutive blue chips will either have a non-blue chip at the top (and what below that?) or will have a blue chip at the top and a non-blue chip second from the top (and what below that?) Mar 28, 2017 at 19:47
• ? thats what i wrote Blue chip then 3 choices then n-2. not blue chip at the top then n-1 satisfies the remaining result? Mar 28, 2017 at 19:49
• You seem to see what is going on correctly., my point is that looking at this without any context, we'd see a line in which you wrote "B3$a_{n-2}$" and not have any idea what you are meaning to say. Using enough words to make it clear will help. Mar 28, 2017 at 19:51
• i dont know a better format to communicate what im thinking ill try looking at the format of how others have solved reccurence relations then try editing it. Mar 28, 2017 at 19:51
• I think the fact that i have no idea what im doing is adding a layer of confusion to what im supposed to write... im just trying to break it into cases and then use the _____ word i dont know to piece the recurrence together cause the example in my textbook does that. Mar 28, 2017 at 20:00

Define $r_n$ as any sequence ending in red, white, or green.
Define $b_n$ as any sequence ending in blue.
Obviously $a_n = b_n + r_n$, and your reasoning to find $r_n$ and $b_n$ is correct.