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Old 2020-10-15, 12:47   #23
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Quote:
Originally Posted by Uncwilly View Post
3, 4, 9, 8, 9, 9, 8, 9, 9, 8...
I'll state an obvious answer, the digits of the decimal expansion of 34/100 + 989/99900 = 6991/19980.
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Old 2020-10-15, 13:41   #24
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Originally Posted by Dr Sardonicus View Post
I'll state an obvious answer, the digits of the decimal expansion of 34/100 + 989/99900 = 6991/19980.
three (5) + four (4) = nine (4)
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
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Old 2020-10-15, 15:28   #25
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Quote:
Originally Posted by Uncwilly View Post
three (5) + four (4) = nine (4)
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
What about the sequence that begins 1, 3, ...?
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Old 2020-10-15, 19:52   #26
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Originally Posted by Dr Sardonicus View Post
What about the sequence that begins 1, 3, ...?
What about 5, 6 ? There are others 1, 1 or 2, 2 or 3, 3
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Old 2020-10-16, 00:32   #27
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Originally Posted by Uncwilly View Post
What about 5, 6 ? There are others 1, 1 or 2, 2 or 3, 3
I asked you first.
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Old 2020-10-16, 05:10   #28
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I would guess the triangular numbers, at least, it's a possibility.
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Old 2020-10-16, 11:52   #29
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Originally Posted by R2357 View Post
I would guess the triangular numbers, at least, it's a possibility.
No, Uncwilly gave the rule: Add the number of letters in the (English) names for two consecutive digits to get the next digit.

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.
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Old 2020-10-16, 12:04   #30
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Originally Posted by Dr Sardonicus View Post
No, Uncwilly gave the rule: Add the number of letters in the (English) names for two consecutive digits to get the next digit.

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.
1 (one) + 3 (three) -> 8 (eight)
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).
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Old 2020-10-16, 12:22   #31
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Originally Posted by Viliam Furik View Post
1 (one) + 3 (three) -> 8 (eight)
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.
Yes, ten is not a decimal digit.

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 not-a-decimal-digit next term.
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Old 2020-10-16, 13:37   #32
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Yes, ten is not a decimal digit.
That is where you make an issue. I said number. You assumed digit.
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Old 2020-10-16, 19:24   #33
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Quote:
Originally Posted by Uncwilly View Post
I said number. You assumed digit.
So you did, and so I did.

Hmm. The question of what starting values give bounded sequences is more interesting than I thought!
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