Sum of odd integers $= x$ How many sums are there that add up to a whole number $x$, and are made of only odd numbers? Each number can be used more than once.
 A: You are asking about partitions into odd parts, which is the same as the number of partitions into distinct parts. The number is given in OEIS A000009 and starts $$1, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 22, 27, 32, 38, 46, 54, 64, 76, 89, 104, 122, 142$$   No simple formula is given.
A: This sounds like a homework question, so I'm not going to answer it, only describe how I would answer it.
First of all, let's get rid of those odd numbers, because they are a bore. It is obvious that if $x$ is odd, there have to be an odd number of terms in the sum, and if $x$ is even, there have to be an even number of terms. So let's each all odd [or even] number $n$ up to and including $x$, and answer the question "How many sums of $n$ odd numbers are there which add up to $x$?".
This is obviously identical to "How many sums of $n$ even numbers are there which add up to $x-n$?".
This is obviously identical to "How many sums of $n$ numbers are there which add up to $\frac{1}{2}(x-n)$?".
You may already know how to answer "How many sums of $n$ numbers are there which add up to $y$?". If not, the way to look at it is this:


*

*Write out the $n$ numbers in unary, with commas in between. For instance, 1+4+2 would be 1,1111,11. 

*Note that there will always be $y+n-1$ symbols in the set; that $n-1$ of them must be commas; and that consecutive commas are allowed, since 0 is an allowable number.

*So there are as many ways of arranging $n$ numbers that add up to $y$ as there are of picking $n-1$ comma positions out of $y+n-1$ possible positions.

*That is, the answer to this part of the question is $\binom{y+n-1}{n-1}$.


Unstitching the argument and going backwards, you need to sum these binomials for all odd $n$ from $1$ to $x$ (if $x$ is odd) or for all even $n$ from $2$ to $x$ (if $x$ is even).
Which is left as an exercise for the reader...
