#### Problem

Let p(n) represent the number of different ways in which n coins can be separated into piles. For example, five coins can separated into piles in exactly seven different ways, so p(5)=7.

OOOOO

OOOO O

OOO OO

OOO O O

OO OO O

OO O O O

O O O O O

Find the least value of n for which p(n) is divisible by one million.

#### Solution

// the pentagonal numbers sequence, i.e. 1, 2, 5, 7, 12, 15, 22, ... let pentagonalNumbers = Seq.unfold (fun state -> Some(state, state+1)) 1 |> Seq.collect (fun n -> [n; -1*n]) |> Seq.map (fun n -> int(0.5 * double(n) * double(3 * n - 1))) // the coefficients sequence, i.e. +, +, -, -, +, +, -, -, ... let pCoefficients = Seq.unfold (fun state -> Some(state, -1*state)) 1 |> Seq.collect (fun n -> [n; n]) // cache results to improve performance let mutable cache = Array.init 100000 (fun n -> if n = 0 then 1I else 0I) // define the function p using the pentagonal numbers let rec p k = if cache.[k] <> 0I then cache.[k] else let pSeq = pentagonalNumbers |> Seq.map (fun n -> k - n) |> Seq.takeWhile (fun n -> n >= 0) |> Seq.map p let pk = pCoefficients |> Seq.zip pSeq |> Seq.sumBy (fun (pk, coe) -> pk * bigint(coe)) cache.[k] <- pk pk let answer = Seq.unfold (fun state -> Some(state, state+1)) 1 |> Seq.filter (fun k -> (p k) % 1000000I = 0I) |> Seq.head

Well, this one took some thought and some time to come up with a solution which runs under a minute! A brute force approach simply wouldn’t have worked here as p(k) becomes very large very quickly (p(200) = 3972999029388…). Thankfully, as the wiki page on partition points out, you can build a recurrence for the function p such that:

*p*(*k*) = *p*(*k* – 1) + *p*(*k* – 2) – *p*(*k* – 5) – *p*(*k* – 7) + *p*(*k* – 12) + *p*(*k* – 15) – *p*(*k* – 22) – …

using pentagonal numbers in the form of

for n running over positive and negative integers (n = 1, –1, 2, –2, 3, –3, …), generating the sequence 1, 2, 5, 7, 12, 15, 22, … The signs in the summation continue to alternate +, +, –, –, +, +, …

Using dynamic programming technique, I’m caching the value of p(k) as they’re generated and in doing so, enabled this solution to run in under 30 seconds.

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**Yan Cui**

I’m an **AWS Serverless Hero** and the author of **Production-Ready Serverless**. I have run production workload at scale in AWS for nearly 10 years and I have been an architect or principal engineer with a variety of industries ranging from banking, e-commerce, sports streaming to mobile gaming. I currently work as an independent consultant focused on AWS and serverless.

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Further reading

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