Problem 44 - Find the smallest pair of pentagonal numbers whose sum and difference is pentagonal.

링크

Pentagonal numbers are generated by the formula, P_n=n(3n1)/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, 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?

python - 1

p = set([x*(3*x-1)/2 for x in range(1, 3000)]) print min(i2 - i1 for i1 in p for i2 in p if i2 - i1 in p and i2 + i1 in p)

python - 2

import math def Pent(n): return n*(3*n-1)/2 def isPent(n): x = int((.5 + math.sqrt(.25+6*n))/3) return n == Pent(x) print min(Pent(b)-Pent(a) for a in range(1, 3000) for b in range(a+1, 3000) if isPent(Pent(b)-Pent(a)) and isPent(Pent(b)+Pent(a)))

1이 2보다 훨씬 빠르다??