Partition function P

Partition function P is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Partition function P
You are encouraged to solve this task according to the task description, using any language you may know.

The Partition Function P, often notated P(n) is the number of solutions where n∈ℤ can be expressed as the sum of a set of positive integers. For example, P(4)=5 because 4=Σ(4)=Σ(3,1)=Σ(2,2)=Σ(2,1,1)=Σ(1,1,1,1).

P(n) can be expressed as the recurrence relation: P(n) = P(n-1) +P(n-2) -P(n-5) -P(n-7) +P(n-12) +P(n-15) -P(n-22) -P(n-26) +P(n-35) +P(n-40)...

The successive numbers in the above equation have the differences: 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8...

This task may be of popular interest because Mathologer made the video, The hardest "What comes next?" (Euler's pentagonal formula), where he asks the programmers among his viewers to calculate P(666). The video has been viewed more than 100,000 times in the first couple of weeks since its release.

In Wolfram Language, this function has been implemented as PartitionsP.

Task

Write a function which returns the value of PartitionsP(n). Solutions can be iterative or recursive. Bonus task: show how long it takes to compute PartitionsP(6666).

References

• The hardest "What comes next?" (Euler's pentagonal formula) The explanatory video by Mathologer that makes this task a popular interest.
• Partition Function P Mathworld entry for the Partition function.
• Partition function (number theory) Wikipedia entry for the Partition function.

Julia

Recursive

<lang Julia>using Memoize

function partDiffDiff(n::Int)::Int

   isodd(n) ? (n+1)÷2 : n+1

end

@memoize function partDiff(n::Int)::Int

   n<2 ? 1 : partDiff(n-1)+partDiffDiff(n-1)

end

@memoize function partitionsP(n::Int)

   T=BigInt
   if n<2
       one(T)
   else
       psum = zero(T)
       for i ∈ 1:n
           pd = partDiff(i)
           if pd>n
               break
           end
           sign = ((i-1)%4)<2 ? 1 : -1
           psum += sign*partitionsP(n-pd)
       end
       psum
   end

end

n=6666 @time println("p($n) = ", partitionsP(n))</lang>

Output:

p(6666) = 193655306161707661080005073394486091998480950338405932486880600467114423441282418165863
  0.405178 seconds (5.34 M allocations: 114.048 MiB, 13.34% gc time)