Why Sequence Questions Are No Good!

[quantyst]

Let’s see if I can show why sequence questions are no good.

The problem with sequence questions is that any answer is in fact a correct answer. Of course, in answering a sequence question, we need to find the most obvious answer. But the issue of “the most obvious answer” is never (or rarely) clarified. Given the sequence 4, 7, 10, 13, 16, 19, 22, the determination of what the next term of the sequence is depends on what mechanism (or formula) generated the first seven terms of the sequence in the first place. And there are infinitely many formulas that can generate the first seven term sof the sequence as given. So, which of these infinitely many formulas did the inquirer have in his/her mind? Of course, for the given sequence, many believe that the simplest mechanism that can generate the sequence is the arithmetic process. But how in general are we supposed to know what the simplest mechanism that generates any given sequence is? The issue is: what does "simplest" mean anyway?

To bring home the problem with sequence questions, let me give you three questions and see if you can answer them:

(1) Find the next four terms in the following sequence:

252, 140, 70, 30, 10, 2, ?, ?, ?, ?



(2) Find the next term in the following sequence:

3, 2, 3, 3, 5, 7, 3, 5, 11, 7, ?



I really would have failed easily answering either one of the preceding questions? I don’t like them unless certain sufficient clues are given along with the question. And here’s one of my favorite (really??) sequence questions:

(3) Find the next three terms of the following sequence:

0, 0, 0, 0, ?, ?, ?



Is there a mensa who'd like to crack it open? Good Luck!

 

[rwinston]

have you just finished reading The Black Swan, by any chance? ;-)

 

[Aditya Chitral]

I think for the third one
the ans is 1,1,1

I will have to give the rest of them a thought.

 

[Aditya Chitral]

For 252,140,70,30,10,2.

the next four terms: 28,0,-10,-10

Am I thinking correct ?

 

[Aditya Chitral]

For 3, 2, 3, 3, 5, 7, 3, 5, 11, 7, ?

Is the answer 5 ?

Well all the answers have an idiosyncratic logic. Hope it works

 

[quantyst]

For 3, 2, 3, 3, 5, 7, 3, 5, 11, 7, ?

Is the answer 5 ?

Well all the answers have an idiosyncratic logic. Hope it works

So, what formula did you use to get the answer 5?

I will tell you my formula after you tell me yours.

BTW, 5 is the correct answer according to my quaint formula, which is basically of the form f(n) -- which denotes the n-th term of the sequence. f(n) is uniquely specified as soon as n is known. Your formula or thinking, whatever it is -- which need not necessarily be in a closed from -- should uniquely determine what the n-th term is.

I am always amazed how someone comes up with the correct answer given the my perceived improbability of doing so.

So, what is your formula?

 

[Aditya Chitral]

n : 1 2 3 4 5 6 7 8 9 10 11 12

f(n) : 3, 2, 3, 3, 5, 7, 3, 5, 11, 7, ?

Here it goes,

for n=1 -- f(n) + f(n+1) = 5, so the 5th number from nth digit is f(n)th prime number
here f(n) = 3
f(n+1) = 2
5th number in sequence will be the 3rd prime number
therefore for n=1 - f(n) = 5

if f(n) is even then (f(n)+f(n+1))th digit from (n+1) is f(n)th prime number
if f(n) is even then (f(n)+f(n+1))th digit from (n) is f(n)th prime number

I know this must be the most weird logic you may have come across, but it worked

what about the other two ??

 

[quantyst]

Thank you for responding.

But I am at a loss. It seems your answer lacks consistency or, at a minimum, clarity. Have you noticed the last two 'if' statements you made? In both you say "if f(n) is even ...", followed by two inconsistent consequents. Just take a look, and you'll see what I mean.

Anyway, it is nonetheless remarkable that you've gone through such a complex thought process. Good for you!

Now, it's only fair that at this time I tell you how I developed the sequence.

Let the function G(x) denote the greatest prime factor of the natural number x, where x>1.

Let p[1], p[2], p[3], .... denote the standard sequence of prime numbers.

Here's my f(n): f(n) = G( ((p[n])^2) - 1 ) = G( (p[n] - 1)*(p[n] + 1) ).

And it is notable that you realized the prime numbers were somehow involved. Good!

Your answer to the third is as good as anyone's. My answer could be anything. But, how about just zeros?

Your answer to the first question is different from mine. I'll give you a clue as to what kind of formula I've used to generate it: polynomial in n of degree four. I'm sure you can figure it out with this clue in hand.

 

[Aditya Chitral]

Oh I am sorry,

that was a typo,

the second one was "odd".

 

[quantyst]

Hi Aditya,

Your answer, pasted below, (even after imagining the second "even" as "odd") remains unclear!

"
if f(n) is even then (f(n)+f(n+1))th digit from (n+1) is f(n)th prime number
if f(n) is even then (f(n)+f(n+1))th digit from (n) is f(n)th prime number
"

In the first line, what do you mean by "... digit from (n+1) is ..."?

What does (n+1) -- which is a natural number -- have to do with the terms of the sequence itself? Remember, you can refer to the terms of the sequence either by writing 'f(n)' or using ordinal language like 'the n-th term' or 'the (n+1)-th term'. You also need to observe that the only time f(n) is even is when n=2. In all other cases, f(n) is odd -- at least, going by the starting terms of the sequence as given. At any rate, it is best to be absolutely crispy clear in defining a function, no matter how you formulate it. It should be so clear that any one with bare understanding of mathematical symbolism can apply your formula to get the same answers as you do. This advice is one I give to myself every time before I offer it to anyone else. Unfortunately, I fail to live up to it every now and then. But, we all try. Don't we?

BTW, did you use my clue to come up with the formula for the first sequence? So, what are the last four missing terms?