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Project Euler 44: Pentagonal numbers

Armaan Nougai

Problem statement

The nth term of the pentagonal sequence is defined as:

 Pn = n*(3*n - 1)/2

Find a pair of two pentagonal numbers Pj and Pk, such that:

 Pk + Pj = S is a pentagonal 
 number
 Pk - Pj = D is a pentagonal 
 number and is minimum too.

What is the value of D?

Question analysis
Question analysis

Solution approach

We will run two loops:

  • an outer loop over pentagonal sequence for the value of P(k).
  • an inner loop over pentagonal sequence less than P(k) for P(j).

Then, we will calculate and check if D and S are pentagonal. To check this, we will inverse the pentagonal function.

The moment we get a pair for which D and S are pentagonal, we’ll break the loop and return D. This is because the first pentagonal D is also the minimum one.

Inverse of pentagonal function
Inverse of pentagonal function

Solution code

"""
Coded by - Armaan Nougai
"""

from math import sqrt

def is_pentagonal(n): return (1+sqrt(1+24*n))%6==0

i=0
while True:
    i+=1
    k = i*(3*i-1)//2
    for v in range(1,i):
        j = v*(3*v-1)//2
        if is_pentagonal(k-j) and is_pentagonal(k+j) :
            print(k-j)
            break
    else:
        continue
    break

A question to address

Why is the first value of D (i.e., 5482660) the minimum possible value of D?

On analyzing the pentagonal sequence:

1. P(2)-P(1) = 4

2. P(3)-P(2) = 7

3. P(4)-P(3) = 10

Generalizing:
P(n)-P(n-1) < P(n+1)-P(n) < P(n+2)-P(n+1) ...

This means the difference between adjacent terms is increasing.

And also, for the nth term, the minimum D, even without the condition, will be:

P(n)-P(n-1)

Now, after getting the first value of D, we will continue the search for the lesser value of D until we find an adjacent pair that satisfies the condition and whose D is greater than the first value of D.

This is because, after this point, the D of every pair will be greater than the first value of D.

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