Solution to problem 44

src/Python/Problem044.py

@@ -0,0 +1,48 @@
+#!/usr/bin/env python3
+"""
+Created on 30 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 44 of Project Euler
+https://projecteuler.net/problem=44
+"""
+
+from itertools import combinations
+from operator import add, sub
+from utils import timeit
+
+def pentagonal(n):
+    return int(n*(3*n-1)/2)
+
+@timeit("Problem 44")
+def compute():
+    """
+    Pentagonal numbers are generated by the formula, Pn=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 P4 + P7 = 22 + 70 = 92 = P8. However, their
+    difference, 70 − 22 = 48, is not pentagonal.
+
+    Find the pair of pentagonal numbers, Pj and Pk, for which their
+    sum and difference are pentagonal and D = |Pk − Pj| is minimised.
+    
+    What is the value of D?
+    """
+    dif = 0
+    pentagonal_list = set(pentagonal(n) for n in range(1,2500))
+    pairs = combinations(pentagonal_list, 2)
+    for p in pairs:
+        if add(*p) in pentagonal_list and abs(sub(*p)) in pentagonal_list:
+            dif = (abs(sub(*p)))
+            # the first one found would be the smallest
+            break
+                
+    return dif
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 44: {compute()}")