We have 2 choices for a coin of a particular denomination, either i) to include, or ii) to exclude.
If we are at coins[n-1], we can take as many instances of that coin ( unbounded inclusion ) i.e count(coins, n, sum – coins[n-1] ); then we move to coins[n-2]. 
After moving to coins[n-2], we can’t move back and can’t make choices for coins[n-1] i.e count(coins, n-1, sum). 
Fin...

diff = no of ways when color of last
two posts is different
same = no of ways when color of last 
two posts is same
total ways = diff + same

for n = 1
diff = k, same = 0
total = k

for n = 2
diff = k * (k-1) //k choices for
first post, k-1 for next
same = k //k choices for common 
color of two posts
total = k + k * (k-1)

for n = 3