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

 Digit Sequence (Posted on 2005-06-28)
Find the rule and continue this sequence of digits:

1235813941213533488671216141383377554...

Is there a limit to the number of times in a row a digit can appear?

 See The Solution Submitted by Jer Rating: 3.6250 (8 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): Theoretically (growing vs. shrinking patterns) | Comment 11 of 15 |
(In reply to re: Theoretically (growing vs. shrinking patterns) by Charlie)

As Charlie points out, there is a string of 9 1's at location 20560, therefore there will be a string of 27 1's sometime later.

Based on the pattern shown in an earlier response, I am led to conclude that there is no limit to the number of times the digits 1, 2, 4, 5, 7, and 8 can occur in a row.

What about 0, 3, 6, and 9?

4 3's makes 3 3's:

3333
666
1212
333

5 3's makes 5 3's:

33333
6666
121212
33333

and 6 3's gives 7 3's:

333333
66666
12121212
3333333

So if 6 or more 3's appear together there is no limit to the number of 3's or 6's.

3 9's make 3 more 9's and 4 9's make 5 9's:

9999
181818
99999

6 3's appear together in Charlie's latest listing, and 4 9's appear together so the only digit with a limit is the digit 0.

Any number (n) of 0's together creates (n-1) 0's, so to get an unlimited number of 0's together you would have to start with an an infinite number.  There are no 2 digit combinations that create 2 or more 0's together.

 Posted by Erik O. on 2005-06-28 20:56:58

 Search: Search body:
Forums (0)