''' How many stones (maximum) can you fit in for given cost? 
    -> nmax(cost)
    Obviously every pile has to give this same cost, that is 
     -> nmax(cost-1^2) + nmax(cost-2^2)+...
'''

from functools import cache
from itertools import count

@cache
def nmax(cost):
    if cost < 4: return 1
    return sum(nmax(cost-pile**2) 
		                      for pile in range(1,int(math.sqrt(cost))+1))
            
def towerBreakers(n):
    for cost in count():
        if nmax(cost)>=n: return cost