4 ms·

What is the next term in the sequence [1, 1, 28, 2, 8, 6, 992, 1, 3, 2, 16256, 2, 16, 16, 523264]? The answer is 24, if anyone was wondering. Coming to this a

by droffel 11y ago

What is the next term in the sequence [1, 1, 28, 2, 8, 6, 992, 1, 3, 2, 16256, 2, 16, 16, 523264]? The answer is 24, if anyone was wondering. Coming to this answer as a "logical conclusion" requires knowing the context of the question, that is, the reader is assumed to have (or somehow discover) the knowledge of how many differential structures there are in N dimensions, for an N starting at 5, and going up to 19. While the paper is fascinating, I doubt that any sane person would think that this is any kind of a realistic "IQ" test question. The title feels like clickbait, and even more so since this "question" wasn't even the topic of the paper at all. Edit: The title has been altered to no longer be clickbait. You can disregard that part of my comment.

cottonseed 11y ago
I like this variant: Complete this sequence: 1, 1, 1, ?, 1, 1, ... If you said infinity, you're right! Number of exotic differentiable structures on R^n. https://en.wikipedia.org/wiki/Exotic_R4 https://en.wikipedia.org/wiki/Exotic_R4
conistonwater 11y ago
Here is OEIS, a general service for solving these types of puzzles and much more: http://oeis.org/search?q=1%2C+1%2C+28%2C+2%2C+8%2C+6%2C+992%2C+1%2C+3%2C+2%2C+16256%2C+2%2C+16%2C+16%2C+523264 http://oeis.org/search?q=1%2C+1%2C+28%2C+2%2C+8%2C+6%2C+992%... You can always just look it up. Also, correction: this is a number of classes of oriented differentiable structures on N-spheres, not just "in N dimensions".
- wolfgke 11y ago
  When I look at this sequence I see that for from 7 on, there seem nearly always exist more than one class of oriented differentiable structures on N-spheres, except for (up to 63) for 12 and 61. Does anybody know what makes the dimensions 12 and 61 so special here?
  - GFK_of_xmaspast 11y ago
    That looks to be addressed here: http://www.ams.org/notices/201106/rtx110600804p.pdf http://www.ams.org/notices/201106/rtx110600804p.pdf
- evanpw 11y ago
  According to this paper (http://arxiv.org/pdf/1601.02184v1.pdf http://arxiv.org/pdf/1601.02184v1.pdf), the entry for n=56 should also be one. Apparently, these values come from a computer-assisted calculation from 1990, and the error wasn't discovered until 2014.
dang 11y ago
Ok, we took out the bit about an IQ question and replaced it with other subtitle. If Gowers says someone is a giant in modern mathematics, that's no clickbait.
- jacobolus 11y ago
  Note: this appreciation is from 2011, when Milnor won the Abel Prize. It’s safe to say that anyone who wins an Abel Prize is a giant in mathematics.
- MakingBreakfast 11y ago
  For the sequence 1,2,5,14,41,122, I didn't notice that you gained each new term by multiplying by three and then subtracting one. However, the pattern I did notice is each additional term is obtained by adding 3^(n-1) where n is the term number. 1 + 3^0 = 2. 2 + 3^1 = 5. 5 + 3^2 = 14. Since these two methods provide the same result, they are probably related somehow, but I can't seem to figure it out.
  - waterhouse 11y ago
    Let f(n) = the nth term of the sequence (where f(1) = 1, f(2) = 2, f(3) = 5, etc). You're saying that f(n) = 3^(n-1) + f(n-1). It also happens that f(n) = 3 * f(n-1) - 1. If we started this sequence with a different value, like 1000, these rules would yield totally different sequences; but it happens that our choice of starting value means these rules are equivalent. Why is this so? It will help to observe that f(n) = (3^(n-1) + 1)/2. Let's check if your rule is then correct: f(n) ?= 3^(n-1) + f(n-1) <=> (plug in observed formula for f) (3^(n-1) + 1)/2 ?= 3^(n-2) + (3^(n-2) + 1)/2 <=> (multiply by 2) 3^(n-1) + 1 ?= 2*3^(n-2) + 3^(n-2) + 1 <=> (subtract 1) 3^(n-1) ?= 2*3^(n-2) + 3^(n-2) <=> (add up multiples of 3^(n-2) on right) 3^(n-1) ?= 3*3^(n-2) <=> (identity) T And then let's check the other rule. f(n) ?= 3*f(n-1) - 1 <=> (plug in formula for f) (3^(n-1) + 1)/2 ?= 3*(3^(n-2) + 1)/2 - 2 <=> (multiply by 2, distribute 3 on right) 3^(n-1) + 1 ?= 3*3^(n-2) + 3*1 - 2 <=> (simplify right) 3^(n-1) + 1 ?= 3^(n-1) + 1 <=> (identity) T (I'm using approximately the proof format described in EWD 1300: https://www.cs.utexas.edu/users/EWD/transcriptions/EWD13xx/EWD1300.html https://www.cs.utexas.edu/users/EWD/transcriptions/EWD13xx/E...) The totally formal thing to do, rather than "observing" the formula, would be to prove by induction that the formula was correct (i.e. that the given rule, "n -> 3n - 1" with n starting at 1, yields it). How would one come up with the formula in the first place? Generally, if your rule involves multiplying by k every time, then the nth term will probably have k^n in it--though if you started with, say, n = 1/2, you wouldn't get anywhere. Is there a totally generic, plug-and-chug procedure that would yield a formula like this? Approximately, yes. For now, I'll just link to the Fibonacci series: https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form
Practicality 11y ago
Right. IQ is generally supposed to measure reasoning and problem solving abilities. While some knowledge is a prerequisite, this "test" would simply be testing your familiarity with a particular set of knowledge. Of course, if you taught students this knowledge first, then having them figure it out is a good test of problem solving ability.
Outdoorsman 11y ago
IQ tests attempt to measure how many "contexts", human "constructs" across various disciplines, a person is familiar with and fluidly capable in... The fewer the contexts...well, I'll leave that to individual interpretation....

by droffel 11y ago

cottonseed 11y ago

I like this variant: Complete this sequence: 1, 1, 1, ?, 1, 1, ... If you said infinity, you're right! Number of exotic differentiable structures on R^n. https://en.wikipedia.org/wiki/Exotic_R4 https://en.wikipedia.org/wiki/Exotic_R4

conistonwater 11y ago

Here is OEIS, a general service for solving these types of puzzles and much more: http://oeis.org/search?q=1%2C+1%2C+28%2C+2%2C+8%2C+6%2C+992%2C+1%2C+3%2C+2%2C+16256%2C+2%2C+16%2C+16%2C+523264 http://oeis.org/search?q=1%2C+1%2C+28%2C+2%2C+8%2C+6%2C+992%... You can always just look it up. Also, correction: this is a number of classes of oriented differentiable structures on N-spheres, not just "in N dimensions".

wolfgke 11y ago
When I look at this sequence I see that for from 7 on, there seem nearly always exist more than one class of oriented differentiable structures on N-spheres, except for (up to 63) for 12 and 61. Does anybody know what makes the dimensions 12 and 61 so special here?
- GFK_of_xmaspast 11y ago
  That looks to be addressed here: http://www.ams.org/notices/201106/rtx110600804p.pdf http://www.ams.org/notices/201106/rtx110600804p.pdf
evanpw 11y ago
According to this paper (http://arxiv.org/pdf/1601.02184v1.pdf http://arxiv.org/pdf/1601.02184v1.pdf), the entry for n=56 should also be one. Apparently, these values come from a computer-assisted calculation from 1990, and the error wasn't discovered until 2014.

dang 11y ago

Ok, we took out the bit about an IQ question and replaced it with other subtitle. If Gowers says someone is a giant in modern mathematics, that's no clickbait.

jacobolus 11y ago
Note: this appreciation is from 2011, when Milnor won the Abel Prize. It’s safe to say that anyone who wins an Abel Prize is a giant in mathematics.
MakingBreakfast 11y ago
For the sequence 1,2,5,14,41,122, I didn't notice that you gained each new term by multiplying by three and then subtracting one. However, the pattern I did notice is each additional term is obtained by adding 3^(n-1) where n is the term number. 1 + 3^0 = 2. 2 + 3^1 = 5. 5 + 3^2 = 14. Since these two methods provide the same result, they are probably related somehow, but I can't seem to figure it out.
- waterhouse 11y ago
  Let f(n) = the nth term of the sequence (where f(1) = 1, f(2) = 2, f(3) = 5, etc). You're saying that f(n) = 3^(n-1) + f(n-1). It also happens that f(n) = 3 * f(n-1) - 1. If we started this sequence with a different value, like 1000, these rules would yield totally different sequences; but it happens that our choice of starting value means these rules are equivalent. Why is this so? It will help to observe that f(n) = (3^(n-1) + 1)/2. Let's check if your rule is then correct: f(n) ?= 3^(n-1) + f(n-1) <=> (plug in observed formula for f) (3^(n-1) + 1)/2 ?= 3^(n-2) + (3^(n-2) + 1)/2 <=> (multiply by 2) 3^(n-1) + 1 ?= 2*3^(n-2) + 3^(n-2) + 1 <=> (subtract 1) 3^(n-1) ?= 2*3^(n-2) + 3^(n-2) <=> (add up multiples of 3^(n-2) on right) 3^(n-1) ?= 3*3^(n-2) <=> (identity) T And then let's check the other rule. f(n) ?= 3*f(n-1) - 1 <=> (plug in formula for f) (3^(n-1) + 1)/2 ?= 3*(3^(n-2) + 1)/2 - 2 <=> (multiply by 2, distribute 3 on right) 3^(n-1) + 1 ?= 3*3^(n-2) + 3*1 - 2 <=> (simplify right) 3^(n-1) + 1 ?= 3^(n-1) + 1 <=> (identity) T (I'm using approximately the proof format described in EWD 1300: https://www.cs.utexas.edu/users/EWD/transcriptions/EWD13xx/EWD1300.html https://www.cs.utexas.edu/users/EWD/transcriptions/EWD13xx/E...) The totally formal thing to do, rather than "observing" the formula, would be to prove by induction that the formula was correct (i.e. that the given rule, "n -> 3n - 1" with n starting at 1, yields it). How would one come up with the formula in the first place? Generally, if your rule involves multiplying by k every time, then the nth term will probably have k^n in it--though if you started with, say, n = 1/2, you wouldn't get anywhere. Is there a totally generic, plug-and-chug procedure that would yield a formula like this? Approximately, yes. For now, I'll just link to the Fibonacci series: https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form

Practicality 11y ago

Right. IQ is generally supposed to measure reasoning and problem solving abilities. While some knowledge is a prerequisite, this "test" would simply be testing your familiarity with a particular set of knowledge. Of course, if you taught students this knowledge first, then having them figure it out is a good test of problem solving ability.

Outdoorsman 11y ago

IQ tests attempt to measure how many "contexts", human "constructs" across various disciplines, a person is familiar with and fluidly capable in... The fewer the contexts...well, I'll leave that to individual interpretation....