TITLE: Partitions into parts from an arithmetic progresion
QUESTION [12 upvotes]: Fix an arithmetic progression $R=(a, a+m, a+2m, \ldots)$, and assume that $gcd(a,m)=1$.  Define $q_R(n)$ as the following coefficients:
$$\prod_{i=0}^\infty (1+ t^{a+mi}) = \sum_{n=0}^\infty q_R(n) t^n
$$
In other words, $q_R(n)$ is number integer partitions of $n$ into distinct parts from $R$.  
Problem 1.  Prove that $q_R(n)$ are increasing for $\ n\ge n(a,m)$ large enough.  
I first assumed this is either standard, well known, or easily follows from the existing results.  Now I am less sure.  My literature search gives only papers like this (A. Tripathi, "Coin exchange problem for arithmetic progressions"). 
Note that for $a=m=1$, we get the usual partitions into distinct parts and the claim follows from Euler's theorem that they are equinumerous with partitions into odd parts. 
More generally, I need to prove that all finite differences are positive for large enough $n$.  Formally, define
$$(t-1)^r \prod_{i=0}^\infty (1+ t^{a+mi}) = \sum_{n=0}^\infty q_R(n,r) t^n
$$
Problem 2.  For every $r\ge 1$, prove that $q_R(n,r)>0$ for $\ n\geq n(a,m,r)$ large enough.

REPLY [4 votes]: [This is more of a comment than an answer, but I lack the reputation.] For partitions with repetitions allowed, the analogue of your Problem #2 is solved by a very general theorem of Bateman and Erdos:
http://www.renyi.hu/~p_erdos/1956-05.pdf
Let $A$ be an arbitrary set of natural numbers. For each nonnegative integer $k$, define $p_k(n)$ so that
$$ \sum_{n=0}^{\infty} p_k(n) X^n = (1-X)^k \prod_{a \in A} (1-X^a)^{-1}. $$
They show that $p_k(n)$ is positive for all sufficiently large $n$ if and only if the following holds: There are more than $k$ elements in $A$, and if we remove an arbitrary subset of $k$ elements of $A$, the remaining elements have greatest common divisor $1$.
Unfortunately they remark that the problem of partitions into distinct parts is much harder and refer to the paper of Roth and Szekeres already mentioned in Gjergji's answer.