# Output from increasing numbers of workers

Yesterday, a question was posted that went along the lines of:

Suppose we have a number of workers, $N$. These workers work at a constant rate of $R$ widgets per unit time. ($R > 0$) We also have a fixed time interval, $I > 0$ (in units of time), such that after $I$ time units, we add another worker.

Is there a closed form for the number of widgets produced, $W$, after $t$ time units pass? ($t$ is not necessarily a multiple of $I$.)

Unfortunately, this question was deleted (by the author) shortly following its posting. (I don't know the circumstances of why the question was deleted. If it's an open contest problem or something, I can remove this question as well--just let me know.)

I solved the problem given above after looking at the graph of output versus input. However, my solution relies on the time interval $I$ being constant. (See my work below.)

My question is, if $I$ were to vary by some rule, could a closed form be found? For example, could a solution be found if we were to add one worker every $F_n$ time-units? (where $F_n$ is the $n$th Fibonacci number)

My work for the old question:

Essentially, at $t = kI$, $k\in\mathbb{N}$, we have: $$W = \left(\frac{k(k+1)}{2}\right)IR = \left(\frac{\frac{t}{I}\left(\frac{t}{I}+1\right)}{2}\right)IR$$ This is because our "overall rate" (rate for all the workers) is $R$ for the first $I$ time-units, $2R$ for the next $I$ time-units, etc. The total output for the first $k$ intervals is the sum of these overall rates times $I$. This is equivalent to the sum of the first $k$ natural numbers times $IR$

At $t$ between multiples of $I$, we simply use the amount at the previous multiple of $I$ and add to it the amount of widgets produced since then: $$W = \left(\frac{\left\lfloor\frac{t}{I}\right\rfloor\left(\left\lfloor\frac{t}{I}\right\rfloor+1\right)}{2}\right)IR + \left(\left\lfloor\frac{t}{I}\right\rfloor + 1 \right)\left(t-\left\lfloor\frac{t}{I}\right\rfloor I\right)R$$

(I haven't paid attention to the units of the above expressions. They work out numerically, but not with respect to their units.)

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