Find a recurrence relation and give initial conditions for the number of words of length $n$ that do not contain two consecutive vowels.

I'm trying to find a recurrence relation for the number of words of length $$n$$ that do NOT contain two consecutive vowels.

I'm trying to relate the problem to a similar one, of bit strings (Say, a relation for the number of bit strings that do not contain four consecutive 0's), but I'm starting to think I don't understand that either.

My though process:

Assume $$a_n$$ represents the number of words that do not contain consecutive vowels. If there are 26 letters in the English alphabet and 5 vowels (not including the letter "y"), then there are $$26^n$$ total words. When researching online, I find that similar problems lead people to using $$a_{n-1}$$, but I don't understand where that comes from. Can anyone offer me some insight? Thank you.

• Oops, typo. Thank you. – help_me Feb 27 '20 at 22:11
• You could look at this question, which is the same except for the number of characters in each group. – Ross Millikan Feb 27 '20 at 22:12

Think as follows: A sequence of length $$n$$ that doesn't end in a vowel can be followed by a non-vowel or a vowel followed by a non-vowel. This gives a recurrence (How many of those are of length 0? 1?). Solve that one, and consider the case where it ends in vowel.