"""Thi is Problem 45 of ProjectEuler
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ...
Pentagonal Pn=n(3n-1)/2 1, 5, 12, 22, 35, ...
Hexagonal Hn=n(2n-1) 1, 6, 15, 28, 45, ...
It can be verified that T285 = P165 = H143 = 40755.
Find the next triangle number that is also pentagonal and hexagonal."""
__author__ = "Andrew Phoenix"
import math
import eulib
#This Method was intended to search along the line of hex numbers
#and break out when it found something that was also a tri and a
#pent. I think it's a little processor crazy.
for x in range (144,1000000):
H=eulib.genhex(x)
print H,
if eulib.ispent(H)==True:
if eulib.istri(H)==True:
print H, 'is the answer. It is H of', x
break
"""This is Problem 44 of ProjectEuler
Pentagonal numbers are generated with this formula:
Pentagonal Pn=n(3n-1)/2 1, 5, 12, 22, 35, ...
Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference is pentagonal and D = |Pk Pj| is minimised; what is the value of D?
"""
__author__ = "Andrew Phoenix"
import math
import eulib
top = 2167
for x in range(1, top):
for y in range(x, top):
if eulib.ispent(math.fabs(eulib.genpent(y) - eulib.genpent(x))):
if eulib.ispent((eulib.genpent(y) + eulib.genpent(x))):
print x, eulib.genpent(x), y, eulib.genpent(
y), eulib.genpent(y) - eulib.genpent(x)
break
"""Thi is Problem 45 of ProjectEuler
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ...
Pentagonal Pn=n(3n-1)/2 1, 5, 12, 22, 35, ...
Hexagonal Hn=n(2n-1) 1, 6, 15, 28, 45, ...
It can be verified that T285 = P165 = H143 = 40755.
Find the next triangle number that is also pentagonal and hexagonal."""
__author__ = "Andrew Phoenix"
import math
import eulib
#This Method was intended to search along the line of hex numbers
#and break out when it found something that was also a tri and a
#pent. I think it's a little processor crazy.
for x in range(144, 1000000):
H = eulib.genhex(x)
print H,
if eulib.ispent(H) == True:
if eulib.istri(H) == True:
print H, 'is the answer. It is H of', x
break
