Yeah. what he said.

(January 20, 2015 at 11:25 am)Exian Wrote: 1, 2, 15, 10, 4, 6, 6, 3, 2...

Does not fit.

I should clarify: The rule is semi-deterministic; each number in the sequence depends on the number before it, but is (generally) not uniquely determined by it.

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(January 20, 2015 at 11:42 am)Jenny A Wrote: A. 1, 6, 11, 6, 11, 16, 21, 16, 21, 26

B. 1, 7, 13, 19, 25, 31, 37, 43, 49

Neither fits. I think I might have made this a little too specific, so what I'm going to do is tell you which of the numbers in the sequences "don't work," that is, don't follow the rule based on the number prior.

In both of these sequences, none of the numbers, aside from the leading "1", follow the rule (that is, a 6 may not follow a 1, a 37 may not follow a 31, etc).

However,
6, 12, 18, 24, 30, 36, 42... does fit.

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(January 20, 2015 at 11:32 am)SteelCurtain Wrote: Every number is an integer.

While this is undeniably true, I'm going to go ahead and aver that the "rule" is a bit more specific than that.

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(January 20, 2015 at 11:55 am)Exian Wrote: Ok, I'm making my second guess because I'm impatient

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The first line is in natural sequential ascending order, the second line is in ascending order, but isn't sequential, the third is neither sequential, not ascending.

My second question is 1, 2, 15, 10, 4, 6, 6, 3, 2...If you answer "No", then my guess is that the rule is no number can occur twice in a row, as that doesn't happen in all three lines. If you answer "Yes" then my answer is the sequence can be anything as long as it starts with "1, 2...", the only other thing in common with the lines.

This is almost foolproof, except "No" could also mean the rule is both: "the sequence has to start with '1, 2..' and the same number can't occur twice in a row."

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I will answer "No, 1, 2, 15, 10, 4, 6, 6, 3, 2..." does not work.

I gave long sequences, but I recommend starting with a couple short simple ones to see if they work. This might be tough but boy are you all going to be proud of yourselves when you figure it out! :-D

(January 20, 2015 at 3:09 am)TRJF Wrote: Yay ^_^

These three sequences all follow the same rule

1, 2, 3, 4, 5, 6, 7, 8, 9...
1, 2, 4, 5, 10, 12, 16, 24, 27...
1, 2, 4, 2, 4, 2, 4, 3, 4...

Like "increasing," the rule is non-deterministic, meaning you can't predict the next number in the sequence from the earlier ones.

You can propose sequences, but you'll have to wait a few hours for feedback; it's sleepy time for me!

I think I figured it out. Is this correct?

7, 14, 16, 24, ...

(January 20, 2015 at 1:47 pm)Irrational Wrote:

(January 20, 2015 at 3:09 am)TRJF Wrote: Yay ^_^

These three sequences all follow the same rule

1, 2, 3, 4, 5, 6, 7, 8, 9...
1, 2, 4, 5, 10, 12, 16, 24, 27...
1, 2, 4, 2, 4, 2, 4, 3, 4...

Like "increasing," the rule is non-deterministic, meaning you can't predict the next number in the sequence from the earlier ones.

You can propose sequences, but you'll have to wait a few hours for feedback; it's sleepy time for me!

I think I figured it out. Is this correct?

7, 14, 16, 24, ...

Yes, that does fit the rule!

(January 20, 2015 at 2:04 pm)TRJF Wrote:

(January 20, 2015 at 1:47 pm)Irrational Wrote: I think I figured it out. Is this correct?

7, 14, 16, 24, ...

Yes, that does fit the rule!

Rule:

For each number in the sequence, there are two factors such that one of them is also a factor of the previous number and the other factor is 1 more or 1 less than the other factor of the previous number.

For example, in sequence 2:

1 = 1 * 1
2 = 1 * 2 (one factor is shared by the previous number, the other factor is different from the other factor of the previous number by 1)
4 = 2 * 2 (2 = 2 and the other 2 is 1 more than 1)
5 = 1 * 5 (compare to 1 * 4 = 4, the previous number)
10 = 2 * 5 (notice 5 is shared by the previous one as a factor and 2 is 1 more than 1)
12 = 2 * 6. It is also equal to 3 * 4
16 = 4 * 4 (compare to 3 * 4). It is also equal to 8 * 2
24 = 8 * 3
27 = 9 * 3

That is *correct*!!! YAY WE HAVE A WINNER!!!

I think there's an easier way to conceptualize it: if x is the nth number in the sequence, then the (n+1)th number differs from x by one of the factors of x. This is equivalent to what you said: you find two factors (y and z) of the nth number, x, and the (n+1)th number has one more or one less of either y or z.

So, for example, the factors of 10 are 1, 2, 5 and 10. Any of the following numbers could follow 10:

0 (10-10), 5 (10-5), 8 (10-2), 9 (10-1), 11 (10+1), 12 (10+2), 15 (10+5), 20 (10+10).

Note that weird things happen when 0 comes up, which I realized after I made the first sequences; I would've had to do some post hoc explanation in that case, and ultimately would have allowed any number to come after 0, because, *technically*, all integers are factors of 0.

(January 20, 2015 at 8:15 pm)Irrational Wrote:

(January 20, 2015 at 2:04 pm)TRJF Wrote: Yes, that does fit the rule!

Rule:

For each number in the sequence, there are two factors such that one of them is also a factor of the previous number and the other factor is 1 more or 1 less than the other factor of the previous number.

For example, in sequence 2:

1 = 1 * 1
2 = 1 * 2 (one factor is shared by the previous number, the other factor is different from the other factor of the previous number by 1)
4 = 2 * 2 (2 = 2 and the other 2 is 1 more than 1)
5 = 1 * 5 (compare to 1 * 4 = 4, the previous number)
10 = 2 * 5 (notice 5 is shared by the previous one as a factor and 2 is 1 more than 1)
12 = 2 * 6. It is also equal to 3 * 4
16 = 4 * 4 (compare to 3 * 4). It is also equal to 8 * 2
24 = 8 * 3
27 = 9 * 3

But 1 followed by 6 or 7 is not allowed. So 1 * 7 both of which share a lowest factor of 1 are not allowed.

This highlights an important rule when figuring out sequence puzzles: think not only about the differences between numbers, but how each difference relates to each of the numbers it's "between", and the sum of the sequence up to that point, and the like.

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(January 20, 2015 at 9:57 pm)Jenny A Wrote:

(January 20, 2015 at 8:15 pm)Irrational Wrote: Rule:

For each number in the sequence, there are two factors such that one of them is also a factor of the previous number and the other factor is 1 more or 1 less than the other factor of the previous number.

For example, in sequence 2:

1 = 1 * 1
2 = 1 * 2 (one factor is shared by the previous number, the other factor is different from the other factor of the previous number by 1)
4 = 2 * 2 (2 = 2 and the other 2 is 1 more than 1)
5 = 1 * 5 (compare to 1 * 4 = 4, the previous number)
10 = 2 * 5 (notice 5 is shared by the previous one as a factor and 2 is 1 more than 1)
12 = 2 * 6. It is also equal to 3 * 4
16 = 4 * 4 (compare to 3 * 4). It is also equal to 8 * 2
24 = 8 * 3
27 = 9 * 3

But 1 followed by 6 or 7 is not allowed. So 1 * 7 both of which share a lowest factor of 1 is are allowed.

I'm not sure I understand... 1 can't be followed by 7 because only 1*1 goes into 1 (if we're staying in integers), so the number that follows it would have to be either 1*(1-1)=0 or 1*(1+1)=2.

Or am I misunderstanding?

That was.... good for me. Was it good for you?

This one has two solutions.

2, 6, 15, 35, 77