# Sequence notation where a has a subscript and a superscript

So from what I've seen of sequences a common format would be something like $a_{n+1} = 2a_n + 5$

I was doing some review for sequences and come across a format that looks something like $a_{n+1}=a_n^2-1$ given $a_1=1$

According to my solution manual, $a_1=1,a_2=0,a_3=-1,a_4=0$

I initially thought that maybe it was just a weird notation and I could write it as $a_n$ but that would just make is a continually decreasing sequences so $a_1=1,a_2=0,a_3=-1,a_4=-2,a_5=-3$ and so on. So this doesnt fit the answer.

I'd like to know how this $a_n^c$ where c is just some number actually works in case it comes up on a test. Any example at all or just an explanation would be great, I can't actually find this notation in the section before the question.

For reference, this is in Briggs, Cochran Calculus Early Transcendentals 2nd edition, 8.1 #20

Edit: Thanks to u/Rob Arthan for this

Since $a_n^2 = a_n \times a_n=(a_n)^2$ this makes the answers from above as

$a_2=(a_1)^2-1=1^2-1=0$

$a_3=(a_2)^2-1=0^2-1=-1$

$a_4=(a_3)^2-1=(-1)^2-1=1-1=0$

and so on and so forth

• $a_n^c$ here means $a_n$ to the $c$-th power. So $a_n^2 = a_n\cdot a_n$, the square of $a_n$. (People do occasionally use superscripts as indexes rather than exponents but that is fairly rare.) – Rob Arthan Nov 19 '17 at 0:04
• That makes a lot of sense and clears this up a lot. Thanks, i'll update my question accordingly and I think that officially makes this solved. Not really sure how to mark is solved without a submitted answer though – Vin Nov 19 '17 at 0:14
• You can also post your own answer – Dylan Nov 19 '17 at 0:45

In the comments, Rob Arthan wrote that "People do occasionally use superscripts as indexes rather than exponents but that is fairly rare." In my opinion, that is reason enough to consider $a_n^c$ as very poor style when ${a_n}^c$ is meant. For extra clarity, you can use parentheses, like so: $(a_n)^c$.
So yeah, $a_{n + 1} = a_n^2 - 1$ should have been written as $a_{n + 1} = {a_n}^2 - 1$ or better yet $a_{n + 1} = (a_n)^2 - 1$.