problem44.py

"""
Pentagon numbers
Project Euler Problem #44
by Muaz Siddiqui

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?
"""
from euler_helpers import timeit
import math

def pentagonal(n):
    return n*(3*n-1)/2 

def is_pentagonal(num):
    if num < 0 :
        return False
    pent = (((24*num) + 1)** 0.5 + 1)/6
    return pent == math.floor(pent)

@timeit
def answer():
    #Let's assume this pair is within the first 10000 pentagonal numbers
    pentagonals = []
    for n in range(1, 10001):
        pentagonals.append(int(pentagonal(n)))
    smallest = 0
    # count up from k and down from j
    k = 1
    while not smallest:
        k += 1
        pent_k = pentagonals[k]
        j = k - 1
        while j > 0:
            pent_j = pentagonals[j]
            first = pent_j + pent_k
            next = pent_k - pent_j
            if is_pentagonal(first) and is_pentagonal(next):
                smallest = next 
                break
            j -= 1
    return smallest