Math Challenge

voice-of-apollo (44)in #math • 7 years ago

The first 6 terms of a sequence are (0,1,2,3,4,5). Each subsequent term is the last digit of the sum of the six previous terms. In other words, the seventh term is 5, since 0+1+2+3+4+5 = 15. So, the eight term is 0, since 1+2+3+4+5+5 = 20.
Can the sub-sequence 13579 occur anywhere in this sequence?

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7 years ago in #math by voice-of-apollo (44)

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jackeown (48) 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?

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voice-of-apollo (44) 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.

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