Project Euler 207: Integer partition equations
Problem Description
For some positive integers k, there exists an integer partition of the form 4^{t} = 2^{t} + k,
where 4^{t}, 2^{t}, and k are all positive integers and t is a real number.
The first two such partitions are 4^{1} = 2^{1} + 2 and 4^{1.5849625…} = 2^{1.5849625…} + 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
Project Euler 207 Solution
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Answer
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Partition is perfect when first element is a power of 2.
I think the original problem statement is intentionally obtuse to make the problem appear more difficult.,
For L = 123456; Smallest m for which P(m) < 1 / 123456 = 6721458093506 [/comments]
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