How to get the N-th word in a sequence Suppose I have an alphabet (e.g. consisting of ABCDEF) and a lexicographic order is defined i.e.

A -> B  ... -> F -> AA -> AB .. -> AF -> BA -> BB -> ... -> BF ... -> FF -> AAA -> ...

Is there a way to check, what the N-th word in the above sequence is? For instance, the first work is "A", the 7th is "AA".
 A: It just looks like a numbering system with base 6 instead of 10.
So I would just use the normal equation for changing bases.
Like $A=0, B=1, C=2, D=3, E=4, F=5$, and use the normal formula to switch the number between decimal system and system based on 6.

Actually, it looks a bit more complicated than that since there is no "zero" in this system.
Define $S_n = \sum_{i=0}^n 6^n$. Given any $N$, let $j$ be the largest integer such that $S_j \leq N$. Let $N_0 = N - S_j$.
Now change $N_0$ into base 6, and fill the digits so that you get $j+1$ many digits. Then switch back to the alphabet system.
For instance, if you want to find 51st word, you would notice that $43 = S_2 < 51$, so $N_0 = 51-43 = 8$. $8$ in base 6 is $12$, so fill the zeroes in front to get $012$. Finally, the word you were looking for would be $ABC$.
A: This is the same question as "How to change a number in decimal system to senary system?". Assume N=a1*6+a2*6^2+a2*6^3+...+ai*6^i(a1,a2,...,ai in {0,1,2,3,4,5}). The jth letter in the word belongs to aj(A if aj=1, B if aj=2 and so on).
A: Since you have six alphabets here, you are going to have to convert the n (term which you want to find, into a base 6 number). 
For example if you want to find the 17th character in that sequence you will have to write something like (17) in base 10 = (25) in base 6.
You then assign each alphabet a number so you will have 
A = 1.
B = 2.
C = 3.
D = 4.
E = 5.
F = 6.
Hence the number 25 in base six is BE. So BE is the 17th number in the sequence.
