Guess the sequence.

[King Author]

Because of that other topic, I've been making up a bunch of numerical sequences just for fun. Try to guess the rule from the first few items in the sequence. Or post your own sequences for others to guess. In case a bunch of people start posting sequences, to make things easier, I suggest naming your sequence Your Name followed by a simple ascending number. See below for what I mean.

Here's a super-easy one. I'll honestly be dumbstruck if the first person who posts doesn't get this.

King Author 1
1, 2, 4, 8, 16, 23, 28, 38, 49, 62, 70, 77, 91, 101, 103, 107

And here's what I'm hoping is a hard one. But not hard because of some ridiculous rule like "add the number of letters in every county in Montana in reverse alphabetical order to get the next number in the sequence." Just oblique. Heh, though with how well I understand math and math-minded people, I wouldn't be surprised if this was the easy one, and KA1 was hard.

King Author 2
1
2
6
24
120
720
5,040
40,320
362,880

[t1mm01994]

Spoiler:

Actually, #2 is much easier than #1. #2 is known as n!, or n*(n-1)*(n-2)*...*2*1, or n! = n-1!*n, 1!=1.
But yeah, after some puzzling, on #1: f(n) = f(n-1) +S(n), where S(n) is the sum of the digits in the last number. aka take the last number and add the number of digits.

[Carlington]

Spoiler:

KA1 follows f(n)=2^n until f(5), where it diverges...not entirely sure how. It looks like we're adding the digits of the previous term together, then adding that sum to the previous term to get the next one.
Is that right?
KA2 is the first n terms of n*(n-1)*(n-2)*...*1, or n!

EDIT: Ninja'd by timm

[mfb]

Finding the rule which was used to generate the numbers can be interesting.
Finding a different rule (which is not overly complicated) can be much more interesting .
Given the length of your lists, I don't think there is a nice different rule, and the easiest rules are already posted.

[King Author]

Timm and Carlington got them right. Although I don't know from formulae, and conceptualized the sequences as such...

KA1: Add the sum of the digits of the previous item.

KA2: Multiply the place in the sequence by the value of the previous item. i.e. to get the sixth number in the sequence, multiply 6 by the previous value (120) to get 720.

Alright, smartypantses; try this one on for size. I think this'll actually be hard to figure out, so I'll give a hint in awhile if no one can make any headway.

King Author 3
1, 3, 4, 7, 11,
13, 14, 17, 21, 23

Finding the rule which was used to generate the numbers can be interesting.
Finding a different rule (which is not overly complicated) can be much more interesting ;).
Given the length of your lists, I don't think there is a nice different rule, and the easiest rules are already posted.

That would be interesting. Er, for clarification, though, for KA1 I just stopped after sixteen items arbitrarily (well, I am a fan of hexadecimals, I tend to round off at 16 as opposed to 10 like normal people), and for KA2 I stopped before item ten because it would've been 3,628,800, which I thought would sort of give the sequence away -- both sequences could go on forever, I only stopped because, well, I'd rather not go on forever :p

Edit: Whoops, screwed up the sequence numero san. Also, I've rearranged the way the sequence is displayed as a hint.

[gmalivuk]

both sequences could go on forever, I only stopped because, well, I'd rather not go on forever :p

Well yeah, but the point is that for any finite sequence, there are infinitely many rules that could have generated that sequence, but would result in literally any desired numbers for the next term.

[King Author]

both sequences could go on forever, I only stopped because, well, I'd rather not go on forever :p

Well yeah, but the point is that for any finite sequence, there are infinitely many rules that could have generated that sequence, but would result in literally any desired numbers for the next term.

Oh! Okay, I didn't understand that. I see now.

Hmm, I guess the first "hint" I should ever give for any sequence, then, should be simply to double the length of the sequence.

[Snark]

Snark 1
1, 4, 8, 13, 20, 26, 36, 46, 55, 65, 75, 87...
Snark 2
1, 4, 8, 13, 20, 26, 35, 45, 54, 63, 74, 85...

These 2 are very, very, similar.

[t1mm01994]

King Author 3

Spoiler:

Taking a stab at 3. Differences are 1,2,3,4,2,1,3,4,2,1,3,4,2,1,3 which appears fairly regular, apart from the beginning. If you leave the first 2 off, I could come up with a gazillion rules.. Or if it started with 1,3, 4. It's just the 2 that doesn't fit in.. Else f(n) = f(n-4) + 10, f(1)=1,f(2)=2, f(3)=4, f(4)=7 would have worked.

[Proginoskes]

King Author 3

Spoiler:

Taking a stab at 3. Differences are 1,2,3,4,2,1,3,4,2,1,3,4,2,1,3 which appears fairly regular, apart from the beginning. If you leave the first 2 off, I could come up with a gazillion rules.. Or if it started with 1,3, 4. It's just the 2 that doesn't fit in.. Else f(n) = f(n-4) + 10, f(1)=1,f(2)=2, f(3)=4, f(4)=7 would have worked.

Maybe the 2 is experimental error?

Try: 1, 3, 7, 12, 18, 26, 35, 45, 56, 69, 83, 98, 114, 131, 150, 170, ...

[King Author]

Sploops, I did fudge the sequence, however, you're way off from how I generated the sequence. As you say, you could come up with any number of rules, but I wanna see if anyone can get mine after my big hint.

King Author 3 revised
1, 3, 4, 7, 11,
13, 14, 17, 21, 23

The way I'm displaying the numbers is the hint.

Snark 1
1, 4, 8, 13, 20, 26, 36, 46, 55, 65, 75, 87...
Snark 2
1, 4, 8, 13, 20, 26, 35, 45, 54, 63, 74, 85...

These 2 are very, very, similar.

Well, the amount added to obtain the next item in the sequence fluctuates up and down, it doesn't steadily increase. What do I know of that does that? Fractals, for one, though I doubt that's it here. There's a few Fibonacci numbers, and a few that are only one or two off, but I can't work out how that might be used. Summing the digits of the individual items doesn't appear to do anything. I don't think the steps between any two items has anything to do with steps earlier or further down the line. Nor do I think they're related to any famous sequence like Pi or anything. I don't think you need one sequence to solve the other, they look like they're both derivable on their own.

I'm stumped. Anyone else wanna take a crack at these before I ask for a hint?

[Snark]

King Author 3

Spoiler:

f(1) = 1, f(2) = 3, f(3) = 4, f(4) = 7, f(n) = f(n - 4) + 10 for n>4. That's the best I've got.

Snark 1 & 2 Hints

Spoiler:

There's more to numbers than numerals.

[King Author]

@Snark: That wasn't the rule I was thinking of. In fact, yours is way more complicated than anything I was thinking of. Granted, I used a pretty-obscure form of math, but nothing your average high-school student would've never heard of. Anyone else, or should I give it up?

[gmalivuk]

Tell us at least this much: does your sequence continue the same way as Snark's formula? As in, 24, 27, 31, 33, 34, 37, and so on?

[Ben-oni]

Snark 1&2:
f(n) = f(n-1) + length(num2str(f(n-1)))
The difference is the formatting used by "num2str".

KA 3:
If you want anyone to keep guessing, you'll have to provide enough terms to make the sequence different from proposed solutions.

[Ankit1010]

King Author 3
1, 3, 4, 7, 11,
13, 14, 17, 21, 23

In the second row, you're keeping the first number fixed and adding it to the 2nd, the 3rd etc..

i.e stick 1 and 3 together - 13
stick 1 and 4 together - 17
..
stick 1 and 11 together - add it to left so 11+10 = 21
then stick 1 and 13 together - 10+13 = 23

Formally though, this is the same as what Snark conjectured : f(n) = f(n-4) + 10 for n > 4

[Snark]

@Ben-oni
Correct! Though you may want to spoiler your results next time.

[Snark]

Snark 3

1, 2, 4, 7, 10, 18, 29...

[Ben-oni]

Snark 3: I think we'll need a few more terms...

Spoiler:

f(1) = 1
f(2) = 2
f(n+1) = f(n) + f(n-1) + (-1)^f(n)

1, 2, 4, 7, 10, 18, 29, 46, 74, 121, 194, ...

Ben-oni 1:
1, 11, 21, 1211, 111221, 312211, 13112221, ...

[Snark]

Snark 3: I think we'll need a few more terms...

Spoiler:

f(1) = 1
f(2) = 2
f(n+1) = f(n) + f(n-1) + (-1)^f(n)

1, 2, 4, 7, 10, 18, 29, 46, 74, 121, 194, ...

You got it.


Ben-oni 1:

Spoiler:

Look-and-say_sequence

[Coding]

Ben-oni 1:
1, 11, 21, 1211, 111221, 312211, 13112221, ...

Spoiler:

Continuation: 1113213211, 31131211131221...

f(0) = 1. Take previous string s of digits. For each sub-string s of repeated digit d, write length of s followed by d. Ex. "1" has one "1," so we write "11."

Coding 1:
0, -1, -1, -2, -1, -4, 0, -6, -1, -5, -2, -10, 4, -12,...

[Sorry, had to edit a couple times.]

[MrDrake]

Coding 1:
0, -1, -1, -2, -1, -4, 0, -6, -1, -5, -2, -10, 4, -12,...

Spoiler:

The sum of the proper divisors of n - n, assuming you start at n=0. Meaning the n=28^th term should be 0 (as it is perfect).

MrDrake 1:

2, 5, 5, 4, 5, 6, 3, 7, 6, 8, 4, 7, 7, ...

[King Author]

AFF (away from forum) for awhile. Back.

I'll just post it. KA3 was (I'm not great with formulae so I'l just use text) "add mod 5 of double the previous term to get the next term." Starting with 1, of course.

1. Doubled is 2. 2 mod 5 is 2, so add 2 to get 3.
3. Doubled is 6. 6 mod 5 is 1, so add 1 to get 4.
4. Doubled is 8. 8 mod 5 is 3, so add 3 to get 7.
etc.

The hint was arranging the numbers five at a time. I was also gonna give another hint by saying something like, "hey gmalivuk, why don't you take a crack at this? As a mod, I'd think you'd have a better chance than anyone else." But that probably wouldn't have helped, because admittedly, doubling the previous term before mod-ing made it basically impossible to guess correctly. If it was just "add mod 5 of the previous term" someone'd've probably got it, especially if I gave the "hey mod" hint.

Ankit1010, that was freaking awesome, though. I probably should've just lied and said "you got it, I didn't understand what Snark posted but the way you did it is the way I did it."

By the by, the way I came up with KA3 was on my own fingers, not by writing an actual formula, so I didn't realize that the additions were regular (+2, +3, +1, +4, +2, +1, +3, +4, etc.), which probably threw everyone off.

[vilidice]

I'm a bit too involved in studying for finals to work any of the above out right now, but I can contribute a few puzzles for now.

Since we haven't seen any sequences other than those of natural numbers yet, and I feel like being creative:
(avoiding tex)

vilidice 1:

(1/2)(1+i)
(1/10)(-1+3i)
(1/34)(-3+5i)
(1/74)(-5+7i)
(1/130)(-7+9i)
(1/202)(-9+11i)

vilidice 2 (probably easier):

-1
i
1/2+ i sqrt(3)/2
(1+i)/sqrt(2)
1/4 + sqrt(5)/4 + i sqrt( 5/8 - sqrt(5)/8)
sqrt(3)/2 + i/2

[Qaanol]

I'm a bit too involved in studying for finals to work any of the above out right now, but I can contribute a few puzzles for now.

Since we haven't seen any sequences other than those of natural numbers yet, and I feel like being creative:
(avoiding tex)

vilidice 1:

(1/2)(1+i)
(1/10)(-1+3i)
(1/34)(-3+5i)
(1/74)(-5+7i)
(1/130)(-7+9i)
(1/202)(-9+11i)

Spoiler:

f(n) = (1-2n + (1+2n)i) / (2 + 8n²)
Where indices start at n=0.

vilidice 2 (probably easier):

-1
i
1/2+ i sqrt(3)/2
(1+i)/sqrt(2)
1/4 + sqrt(5)/4 + i sqrt( 5/8 - sqrt(5)/8)
sqrt(3)/2 + i/2

Spoiler:

Solution to xⁿ⁺¹+1=0 with smallest positive argument.

[Ben-oni]

AFF (away from forum) for awhile. Back.

I'll just post it. KA3 was (I'm not great with formulae so I'l just use text) "add mod 5 of double the previous term to get the next term." Starting with 1, of course.

1. Doubled is 2. 2 mod 5 is 2, so add 2 to get 3.
3. Doubled is 6. 6 mod 5 is 1, so add 1 to get 4.
4. Doubled is 8. 8 mod 5 is 3, so add 3 to get 7.
etc.

The hint was arranging the numbers five at a time. I was also gonna give another hint by saying something like, "hey gmalivuk, why don't you take a crack at this? As a mod, I'd think you'd have a better chance than anyone else." But that probably wouldn't have helped, because admittedly, doubling the previous term before mod-ing made it basically impossible to guess correctly. If it was just "add mod 5 of the previous term" someone'd've probably got it, especially if I gave the "hey mod" hint.

Ankit1010, that was freaking awesome, though. I probably should've just lied and said "you got it, I didn't understand what Snark posted but the way you did it is the way I did it."

By the by, the way I came up with KA3 was on my own fingers, not by writing an actual formula, so I didn't realize that the additions were regular (+2, +3, +1, +4, +2, +1, +3, +4, etc.), which probably threw everyone off.

The way you did it, is, of course, exactly the same as: a_n = a_n-4 + 10.

Let a_n be the sequence in question. Then let 0 ≤ p_n < 5 such that p_n ≡ 2a_n (mod 5). Now, a_n+1 = a_n + p_n. Therefore, a_n+1 ≡ a_n + p_n ≡ 3a_n (mod 5).

Let k_n+1 = 3k_n where k₀ = a₀ = 1. Then p_n, the difference between one term and the next, is dependent only upon k_n, which is: 1, 3, 9, 27, ... Therefore, p_n comes from k_n, being the cycle generated by 3 in the (multiplicative) group Z₅ \ {0}, which starts at 2: 2, 1, 3, 4, ... This, of course, repeats every four terms (since 〈Z₅\{0}, ⋄〉≅〈Z₄, +〉, which is a cyclic group), so a_n = a_n-4 + p_n-4 + p_n-3 + p_n-2 + p_n-1 = a_n-4 + 10.