All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers > Sequences
Alternating sextet (Posted on 2007-03-30) Difficulty: 3 of 5
In the sequence 1, 0, 1, 0, 1, 0, 3, 5... each member after the sixth one equals the units' digit of the sum of the six preceding numbers of the sequence.

Prove that the subsequence 0, 1, 0, 1, 0, 1, will never occur.

No Solution Yet Submitted by e.g.    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution More brute force (spoiler) | Comment 3 of 6 |
I just calculated the sequence which starts with the target subsequence 101010.  (I used excel, my tool of preference).  Interesting enough, it also repeated every 1456 numbers.  It did not include 010101, so this is another way to show that one will not generate the other.

I sure hope somebody comes up with something elegant.

  Posted by Steve Herman on 2007-03-31 00:31:00
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information