'''
Created on 7 feb. 2012

@author: Julien Lengrand-Lambert
DESCRIPTION : Solves problem 45 of Project Euler
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(3(n-1))/2       1, 5, 12, 22, 35, ...
Hexagonal        Hn=n(2(n-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.
'''

def tn(value):
    """
    Returns the triangle number of the given value
    """
    return value * (value + 1) / 2

def pn(value):
    """
    Returns the Pentagonal number of the given value
    """
    return value * ((3 * value) - 1) / 2

def hn(value):
    """
    Returns the Hexagonal number of the given value
    """
    return value * ((2 * value) - 1)

def tripenhex(value):
    """
    Returns True of the triangle, pentagonal and hexagonal number of value are the same
    """
    return (tn(value) == pn(value) == hn(value))

def big_while():
    """
    Returns the value for which tm = pn = ho
    """
    cpt_t = 2
    cpt_p = 2
    cpt_h = 2
    val_t = tn(cpt_t)
    val_p = pn(cpt_p)
    val_h = hn(cpt_h)
    
    res = []
    while len(res) < 2: # to get the two first values
        while not (val_t == val_p == val_h):
            
            cpt_t += 1
            val_t = tn(cpt_t)
            
            # updating val_p
            while val_p < val_t:
                cpt_p += 1
                val_p = pn(cpt_p)
                
            # updating val_h
            while val_h < val_t:
                cpt_h += 1
                val_h = hn(cpt_h)    
        res.append(val_t)
        cpt_t += 1
        val_t = tn(cpt_t)
        
    return res[1] # outputs the second value

if __name__ == '__main__':
    
    print "Answer : %d " % (big_while())