Members of the sequence greater than 1 less than N Suppose N is a positive integer. How many decreasing
integer sequences are there such that members of the
sequence are greater than 1 but less than N… 
I have tried to come up with an expression for this problem.
Do i need to look into prime numbers? excluding 1 and the number makes me think so...
 A: Since the OP has been given a fully-formed answer, I'd like to point out an argument that gets to the final form without passing through a summation.
We can equivalently solve for the number of sequences with members between $1$ and $N-2$ inclusive, by subtracting $1$ from each member.
However, notice that the decreasing condition implies that no two members are the same, and moreover if I tell you the set of members then in fact the sequence is entirely determined. Moreover, every subset of $\{1,2,\dots, N-2\}$ corresponds to a sequence of the desired type, except for the empty set, since we need at least one term.
Therefore, the number of sequences of the desired type is equal to the number of nonempty subsets of $\{1,2,\dots, N-2\}$, which is $2^{N-2}-1$.
A: Let $I_k$ be  the number of such sequences with $k$ terms. For example $I_{N-2}=1.$  Then the number of such sequences is 
$$
I_{N-2}+I_{N-3}+\cdots+I_{1}
$$
It is clear that  $I_{N-3}={N-2 \choose 1},$ $I_{N-3}={N-2 \choose 2}$ and so on. Thus 
$$
I_{N-2}+I_{N-3}+\cdots+I_{1}=\sum_{i=0}^{N-3}{N-2 \choose i}=2^{N-2}-1.
$$ 
