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I recently saw a post about sequences. This made me remember some other post someone had posted here on Math.SE. He did not want answers but wanted general ways to tackle them. I did spend an hour or so on these and yet I could not come up with complete solutions to all of these. That is why I am reposting this with different question. What are solutions to these?

1: $\{4, 2 ,36, 4, 5, 1, 36, 16, 6, 3, 81, \ldots\}$

2: $\{\frac{1}{4},\frac{1}{4},\frac{1}{2},\frac{3}{2},6,\ldots\}$

3: $\{7,9,11,6,11,8,5,13,5,4,15,\ldots\}$

4: $\{2,1,2,3,2,9,9,0,1,1,9,\ldots\}$

5: $\{2, 1, 2, 3, 8, 0, 1, 9, 1, 2, 1,\ldots\}$

6: $\{4,2,5,9,5,11,13,7,16,17,9,\ldots\}$

The first three seem to have solutions:

1 $64,7,2,81,...$

2 $30,180,1260...$

3 $2,3,17,...$

I am looking for solutions for last three.

P.S. I accept (as pointed out in other thread) that you can take any number and generate a formula for sequence such that next term can be chosen as wished. But, I am looking for general terms that are solvable without much ado and with simplest formulas or patterns.

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So when you say "solution" you mean "find the next three or four terms"? Oh, and can you give a link to the past post(s)? –  Gerry Myerson Feb 21 '13 at 22:59
    
yes, i do mean so. Link –  user45099 Feb 22 '13 at 6:51
    
Related to: math.stackexchange.com/questions/291443/… –  minopret Feb 22 '13 at 7:08
    
These sequence problems are among the many that are collected here: m4maths.com/placement-puzzles.php?SOURCE=IBM –  minopret Feb 22 '13 at 7:27
    
It's not related to. It's copy from. I clarified, I want solution to these that the original poster did not want. –  user45099 Feb 22 '13 at 9:47

1 Answer 1

up vote 1 down vote accepted

If you consider it legitimate to use the web, you can solve some by typing them into oeis.org or even Google web search, possibly after multiplying by the least common denominator.

It seems inherent to sequence questions that selecting the "correct" answer can be in part a matter of aesthetics. Without resort to aesthetics, or perhaps Chaitin-Kolmogorov complexity, I think you'd have difficulty to disprove my claim that the "correct" continuation of every finite sequence is 0, 0, 0, 0, ...

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Yes, that's what I mentioned in last paragraph, you can continue anyway. But Google was good for first only. I did use OEIS but the only one I found at OEIS is second which did not take much for me to solve. –  user45099 Feb 22 '13 at 7:01
    
OEIS has the third one. I'll see whether I can do anything with the others. Aha, yes, I can find the previous post that was mentioned. –  minopret Feb 22 '13 at 7:30

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