
A049982 is the number of arithmetic progressions of 2 or more positive integers, strictly increasing with sum n.
When I stumbled on this sequence (and it's brothers and sisters, with various slightly different qualifications), I noticed a complete lack of any formulas or generating functions that help understand the sequence. So I did some amateur investigation on my own.
I started by considering the number of arithmetic progressions of 2 positive integers, strictly increasing with sum n. By convention, I like to start with n=0, so this sequence is 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, ... which has generating function x^{3}/(x^{3}x^{2}x+1) which I will rewrite as x^{3}/(x^{3}xx^{2}+1) for reasons that will become clear later.
The sum of an arithmetic progression of 3 positive integers is always three times its middle term, hence a multiple of 3. This sequence is 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 3, 0, 0, 4, ... which has generating function x^{6}/(x^{6}2x^{3}+1), which I will write as x^{6}/(x^{6}x^{3}x^{3}+1) for reasons that will become clear later.
The sum of an arithmetic progression of 4 positive integers follows a pattern that's a little harder to discern. But if I give you enough terms, I think you can start to pick up the rhythm: 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 2, 0, 2, 0, 2, 0, 2, 0, 3, 0, 2, 0, 3, 0, 3, 0, 3, 0, 3, 0, 4, 0, 3, ... This has generating function x^{10}/(x^{10}x^{6}x^{4}+1)
As you can imagine, I kept going. As the going got tougher, I started inventing little tools to help, such as PuzzleGeneratingFunction.xls, an Excel spreadsheet that guesses the generating function for a given series. (. . . . . . maybe I'll write a page about that spreadsheet some day.)
After a while, I had a little table that shows the
generating function for the sequence of the number of arithmetic progressions of k positive integers, strictly increasing with sum n:
k  Generating Function 
2  x^{3}/(x^{3}xx^{2}+1) 
3  x^{6}/(x^{6}x^{3}x^{3}+1) 
4  x^{10}/(x^{10}x^{6}x^{4}+1) 
5  x^{15}/(x^{15}x^{10}x^{5}+1) 
6  x^{21}/(x^{21}x^{15}x^{6}+1) 
7  x^{28}/(x^{28}x^{21}x^{7}+1) 
8  x^{36}/(x^{36}x^{28}x^{8}+1) 
9  x^{45}/(x^{45}x^{36}x^{9}+1) 
10  x^{55}/(x^{55}x^{45}x^{10}+1) 
11  x^{66}/(x^{66}x^{55}x^{11}+1) 
Now, maybe you can see why I wrote the terms for k=2 and k=3 in such a funny way. In general, the generating function for the sequence of the number of arithmetic progressions of k positive integers, strictly increasing with sum n is:
x^{t(k)}/(x^{t(k)}x^{t(k1)}x^{k}+1), where t(k) is the k'th triangular number
A049982 has generating function x^{3}/(x^{3}xx^{2}+1) + x^{6}/(x^{6}x^{3}x^{3}+1) + x^{10}/(x^{10}x^{6}x^{4}+1) + ... which is the
sum k=2,3,... of x^{t(k)}/(x^{t(k)}x^{t(k1)}x^{k}+1), where t(k) is the k'th triangular number
Term k of this generating function generates the number of arithmetic progressions of k positive integers, strictly increasing with sum n.
A049982  The Online Encyclopedia of Integer Sequences.
See also Recurrence Relation
The webmaster and author of this Math Help site is Graeme McRae.