20201015, 12:47  #23 
Feb 2017
Nowhere
3·1,193 Posts 

20201015, 13:41  #24  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3·5·11·53 Posts 
Quote:
four (4) + nine (4) = eight (5) nine (4) + eight (5) = nine (4) nine (4) + nine (4) = eight (5) and around we go The sum of the letters leads to the next number in sequence. one (3) + two (3) = six (3) loops right away 

20201015, 15:28  #25 
Feb 2017
Nowhere
3·1,193 Posts 
What about the sequence that begins 1, 3, ...?

20201015, 19:52  #26 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3·5·11·53 Posts 

20201016, 00:32  #27 
Feb 2017
Nowhere
3·1,193 Posts 

20201016, 05:10  #28 
"Ruben"
Oct 2020
Nederland
37 Posts 
1, 3
I would guess the triangular numbers, at least, it's a possibility.

20201016, 11:52  #29  
Feb 2017
Nowhere
3·1,193 Posts 
Quote:
This would appear to be problematic when you start with 1, 3. Hence my question. Of the 81 pairs of initial nonzero decimal digits, 36 of them cause the same problem. 

20201016, 12:04  #30  
Jul 2018
Martin, Slovakia
223 Posts 
Quote:
3 (three) + 8 (eight) > 10 (ten) 8 (eight) + 10 (ten) > 8 (eight)... The sequence would continue into infinity, alternating between 8 and 10. I can see is that 3,8,? is a bit problematic, because of the number ten not being a single digit. 1,3,? is not problematic, you can do it, but AFTER that comes trouble. If you extend the sequence rule to a number of letters in the name of the number, it becomes possible to do every starting value (I think). 

20201016, 12:22  #31  
Feb 2017
Nowhere
3·1,193 Posts 
Quote:
Another possibility is to interpret a sum of 10 as 1,0. This is precisely why I asked Uncwilly. It was his rule, so it's his call on what to do when the rule gives a notadecimaldigit next term. 

20201016, 13:37  #32 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
8745_{10} Posts 

20201016, 19:24  #33 
Feb 2017
Nowhere
DFB_{16} Posts 

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