Sort:  
  • Trending
    • [-]
     7 years ago 

    I wrote a little python script to do it for me:

    l = (0,1,2,3,4,5)
hist = []
def getNext(l):
        return (l[1],l[2],l[3],l[4],l[5],int(str(sum(l))[-1]))

for i in range(100000):
        hist.append(l)
        l = getNext(l)
        print(l)
        if l in hist:
                print("DUPLICATE at {}".format(i))
                break
        if l == (1,3,5,7,9):
                print("FOUND IT")

    After 1456 iterations of this loop, we come back to the original 6 terms of the sequence and we still haven't encountered 13579.

    I have no good mathematical answer for why this is true though...which I assume there is?

    $0.03
    • Past Payouts $0.03, 0.00 TRX
    • - Author $0.03, 0.00 TRX
    • - Curators $0.00, 0.00 TRX
    2 votes
    Reply
    [-]
     7 years ago 

    Actually I'm still trying to solve this myself. I really like the code. There could be a sequence of 2x+1 starting with x=0 and going to x=4 that is past 1456 iterations. The way to prove it would be to find a recurring cycle and show that this cannot be part of the cycle. What I'm trying to do right now is see if it has any pattern to it.

    $0.02
    • Past Payouts $0.02, 0.00 TRX
    • - Author $0.01, 0.00 TRX
    • - Curators $0.00, 0.00 TRX
    1 vote
    Reply
    [-]
     7 years ago 

    That's what I'm saying I did...the code finds a cycle at the 1456th iteration

    Loading...