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PROJECT EULER · #0207

Integer Partition Equations

Statement only · SolvedOriginal problem ↗

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 m1 let P(m) be the proportion of such partitions that are perfect with km.
Thus P(6)=1/2.

In the following table are listed some values of P(m).

P(5)=1/1P(10)=1/2P(15)=2/3P(20)=1/2P(25)=1/2P(30)=2/5P(180)=1/4P(185)=3/13

Find the smallest m for which P(m)<1/12345.

Write-up coming later

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