#### Problem

Pentagonal numbers are generated by the formula, P_{n}=n(3n-1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, …

It can be seen that P_{4}+ P_{7}= 22 + 70 = 92 = P_{8}. However, their difference, 70 – 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, P_{j}and P_{k}, for which their sum and difference is pentagonal and D = |P_{k}– P_{j}| is minimised; what is the value of D?

#### Solution

open System.Collections.Generic // define the Pentagonal function P let P n = n*(3*n-1)/2 // use a dictionary as cache let mutable cache = new Dictionary<int, int>() [1..5000] |> List.iter (fun n -> cache.[n] <- P n) // function to check if a number is pentagonal by looking at the cached values let isPentagonal n = cache.ContainsValue n // predicate function to check if Pk and Pj's sum and diff are both pentagonal let predicate (k, j) = let pk = cache.[k] let pj = cache.[j] isPentagonal (pj + pk) && isPentagonal (pk - pj) // the sequence of k, j pairs to check let kjSeq = [1..5000] |> List.collect (fun k -> [1..k-1] |> List.rev |> List.map (fun j -> (k, j))) // get the first pair of k, j whose sum and difference are both pentagonal let (k, j) = kjSeq |> Seq.filter predicate |> Seq.head let answer = (P k) - (P j)

Having experimented with a few alternative implementations which did not use a mutable dictionary and waited impatiently for the code to never return I settled on this solution which caches the first 5000 values in the pentagonal number sequence for quick lookup later on in the solution.

Going into more detail about the solution itself, to ensure |Pk – Pj| is minimised, the value of k and the difference between k and j must be minimised too as the further apart Pk and Pj are in the pentagonal sequence the bigger their difference will be and the bigger k is the bigger Pk will be too.

The rest of the algorithm here is simple, for each value k (between 1 and 5000, being the size of the cache) check for each j where 1 <= j < k, starting from the biggest j, if the sum and difference of Pk and Pj are both pentagonal. The first k and j which satisfy this predicate will have the smallest difference between Pk and Pj.

**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.

Further reading

Here is a complete list of all my posts on serverless and AWS Lambda. In the meantime, here are a few of my most popular blog posts.

- Lambda optimization tip – enable HTTP keep-alive
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- Many faced threats to Serverless security
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- Yubl’s road to Serverless
- AWS Lambda – should you have few monolithic functions or many single-purposed functions?
- AWS Lambda – compare coldstart time with different languages, memory and code sizes
- Guys, we’re doing pagination wrong