Question

length = 0
for n in range(1,101):
    print "Sequence #:", n
    while n != 1:
        print n,
        if n % 2 == 0:
           n = n / 2
           length = length + 1
        else:
            n = (n * 3) + 1
            length = length + 1
        if n == 1:
            print n
            length = length + 1
    print "The sequence above contains", length, "numbers"
    length = 0

My Problem:

The python code above calculates the hailstone sequence for numbers 1 – 100 and displays the length of the sequence afterwards. How can I display the number with the longest sequence and its corresponding length after all calculations are done?

Answer 1

This will keep track of the max length and sequence and display them at the end. The lines marked with a ## are the additions to your original code.

length = 0
max_length = 0 ##

for n in range(1,101):
    print "Sequence #:", n
    seq = [] ##
    while n != 1:
        print n,
        seq.append(n)  ##
        if n % 2 == 0:
           n = n / 2
           length = length + 1
        else:
            n = (n * 3) + 1
            length = length + 1
        if n == 1:
            print n
            seq.append(n) ##
            length = length + 1
    print "The sequence above contains", length, "numbers"

    if length > max_length:  ##
        max_length = length  ##
        max_seq = seq[:]     ##

    length = 0

print 'max length: ', max_length ## same as len(max_seq)
print 'max seq: ', max_seq       ##