Problem 207
Integer partition equations
For some positive integers k, there exists an integer partition of the form 4t = 2t + k,
where 4t, 2t, and k are all positive integers and t is a real number.
The first two such partitions are 41 = 21 + 2 and 41.5849625… = 21.5849625… + 6.
Partitions where t is also an integer are called perfect.
For any m ≥ 1 let P(m) be the proportion of such partitions that are perfect with k ≤ m.
Thus P(6) = 1/2.
In the following table are listed some values of P(m)
P(5) = 1/1
P(10) = 1/2
P(15) = 2/3
P(20) = 1/2
P(25) = 1/2
P(30) = 2/5
…
P(180) = 1/4
P(185) = 3/13
Find the smallest m for which P(m) < 1/12345
整数分拆等式
对于某些正整数k，存在形如4t = 2t + k的整数分拆，
其中4t，2t和k均为正整数，而t为实数。
前两个这样的分拆分别是41 = 21 + 2和41.5849625… = 21.5849625… + 6。
如果t也是整数，这样的分拆称为完美的。
对于任意m ≥ 1，记P(m)是所有k ≤ m的分拆中完美分拆的比例。
因此P(6) = 1/2。
下面的表格列出了P(m)的部分取值：
P(5) = 1/1
P(10) = 1/2
P(15) = 2/3
P(20) = 1/2
P(25) = 1/2
P(30) = 2/5
…
P(180) = 1/4
P(185) = 3/13
求出使得P(m) < 1/12345的最小m值。