Combinatorics is not my thing, so I am kind of struggling with this.
Given positive real numbers $\varepsilon$ (typically small) and $T$ (typically big), how many solutions in $\mathbb{N}_0\times\mathbb{N}_0$(or at least bounds) are there for the inequality
$$n_1\varepsilon +n_2\leq T $$
Without the $\varepsilon$-weight the answer would be (if T is a natural number) $\sum_{k=0}^2\binom{2}{k}\binom{T}{k}$, so I imagine that with small $\varepsilon$ the answer should be something bigger.
I was thinking something like for fixed $n_2$ (which should be between $0$ and $T$) I can count how many $n_1$ satisfies $n_1\leq \frac{T-n_2}{\varepsilon}$. Then I can sum it for $0\leq n_2\leq T$, but it does not seem to work correctly.
It seems to be a standard question in combinatorics, but I could find it anywhere. If anyone has any reference I would be greatfull too.
Answer: $$ \sum_{k=0}^T\left(\left\lfloor\frac{T-k}{\epsilon}\right\rfloor+1\right) $$ Explanation: Count the number of solutions where $n_2=k$, then sum over all possible $k$. If $n_2=k$, we require $n_1\le (T-k)/\epsilon$. For any $x\in \mathbb R_{\ge 0}$, the number of nonnegative integers which are $\le x$ is $\lfloor x\rfloor +1$.