The following iterative sequence is defined for the set of positive integers:
It can be seen that this sequence (starting at 13 and finishing at 1)
contains 10 terms. Although it has not been proved yet (Collatz
Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
Python Code:
def Collatz():
lengthCollatz = 1
count = 1
numb = 0
longestCnumb =1
for i in range(1000000):
count = 1
numb = i+1
while numb != 1:
if numb%2 == 0:
numb = numb/2
else:
numb = 3*numb+1
count +=1
if count > lengthCollatz:
lengthCollatz = count;
longestCnumb = i+1;
print "Longest length of Collatz Sequence is:",lengthCollatz,"and the number is",longestCnumb
Collatz()
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:n → 3n + 1 (n is odd)
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Which starting number, under one million, produces the longest chain?
Python Code:
def Collatz():
lengthCollatz = 1
count = 1
numb = 0
longestCnumb =1
for i in range(1000000):
count = 1
numb = i+1
while numb != 1:
if numb%2 == 0:
numb = numb/2
else:
numb = 3*numb+1
count +=1
if count > lengthCollatz:
lengthCollatz = count;
longestCnumb = i+1;
print "Longest length of Collatz Sequence is:",lengthCollatz,"and the number is",longestCnumb
Collatz()
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