A little puzzle

[Dr Sardonicus]

Originally Posted by Uncwilly

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.

[Uncwilly]

Originally Posted by Dr Sardonicus

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

[Dr Sardonicus]

Originally Posted by Uncwilly

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, ...?

[Uncwilly]

Originally Posted by Dr Sardonicus

What about the sequence that begins 1, 3, ...?

What about 5, 6 ? There are others 1, 1 or 2, 2 or 3, 3

[Dr Sardonicus]

Originally Posted by Uncwilly

What about 5, 6 ? There are others 1, 1 or 2, 2 or 3, 3

I asked you first.

[R2357]

I would guess the triangular numbers, at least, it's a possibility.

[Dr Sardonicus]

Originally Posted by R2357

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.

[Viliam Furik]

Originally Posted by Dr Sardonicus

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).

[Dr Sardonicus]

Originally Posted by Viliam Furik

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.

[Uncwilly]

Originally Posted by Dr Sardonicus

Yes, ten is not a decimal digit.

That is where you make an issue. I said number. You assumed digit.

[Dr Sardonicus]

Originally Posted by Uncwilly

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!