An Infinite Partition for 2016

$\textbf{Generalization:}$ $\frac{a_1}{d_1} + \frac{a_2}{d_2} + ... + \frac{a_k}{d_k} = \frac{k-1}{2}$ (MOSP)

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Note that the following is true $\sum_{n = 0}^{\infty}{X^{a_1 + nd_1}} + ... + \sum_{n = 0}^{\infty}{X^{a_k + nd_k}} = \sum_{n = 0}^{\infty}{X^n}$ which gives us for $|X| < 1$ , $\begin{aligned} \sum_{i = 1}^{k}\frac{X^{a_i}}{1 - X^{d_i}} & = \frac{1}{1-x} \\ \sum_{i = 1}^k\frac{X^{a_i}}{1 + X + ... + X^{d_i - 1}} & = 1 \end{aligned}$ Taking the limit when $X \rightarrow 1^-$ gives us $\boxed{\sum \frac{1}{d_i} = 1.}$

Consider the second equation,

$\sum_{i = 1}^k\frac{X^{a_i}}{1 + X + ... + X^{d_i - 1}} = 1$ taking $\frac{d}{dx}$ of both sides gives us $\sum_{i = 1}^k\frac{a_iX^{a_i - 1}(1 + X + ... + X^{d_i - 1}) - X^{a_i}(1 + 2X + ... + (d_i - 1)X^{d_i - 2})}{(1 + ... + X^{d_i - 1})^2} = 0$ and taking the limit of $X \rightarrow 1^-$ gives us $\begin{aligned} \sum_{i = 1}^k \frac{a_id_i - \frac{(d_i - 1)(d_i)}{2}}{d_i^2} & = 0 \\ \sum_{i = 1}^k \frac{a_i - \frac{d_i - 1}{2}}{d_i} & = 0 \\ \sum{\frac{a_i}{d_i}} & = \frac{k-1}{2} \end{aligned}$ and we are done.