012.py

#!python


problem = """
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The fi
rst ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

 1: 1
 3: 1,3
 6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

https://projecteuler.net/problem=12

OR

Must find the smallest number t, such that
t = n*(n+1)/2
for all n, also satisfying

t = PI(p^m)
is the prime factorization of t.

The number of divisors is D = PI(m+1)

600 = 2^3 * 3 * 5^2 --> (3+1)(1+1)(2+1) = 24 divisors

"""

print(problem)

# first get the list of the first X primes:
import math
top=3000
primeset=[2,3]
m=1
x=len(primeset)
def check_prime(n):
  global x,primeset
  primetest=True
  topn=math.sqrt(n)
  for i in primeset:
    if i > topn:
      break
    if (n%i) == 0:
      primetest=False
  if primetest == True:
    primeset.append(n)
    x=len(primeset)
    # print("found",x,n)
while x < top:
  check_prime((6*m)-1)
  check_prime((6*m)+1)
  m+=1



n = 2
bigtop=50000
dictofmax={'n':0,'Tri':0,'Expos':0}
numofdivs=0

while numofdivs < 500:
  keeptri = int(n * ( n+1 ) /2)
  tri=keeptri
  listofdivs=[]
  while tri > 1:
    for m in primeset:
      if m <= tri:
        if ( tri % m ) == 0:
          listofdivs.append(m)
          tri /= m
  setofdivs=set(listofdivs)
  dicty={}
  numofdivs=1
  for p in setofdivs:
    expo=listofdivs.count(p)
    numofdivs*=(expo+1)
    dicty[p]=expo

  if numofdivs > dictofmax['Expos']:
    dictofmax['n']=n
    dictofmax['Tri']=keeptri
    dictofmax['Expos']=numofdivs
    print(n,keeptri,listofdivs,setofdivs,dicty,numofdivs,dictofmax)

  n+=1

print()
print(keeptri,numofdivs)