Given (2,3) (2, 5) (2, 11) (2, 17), find next three sets

[frappa]

Can you figure out the next three sets?

(2,3) (2, 5) (2, 11) (2, 17)

I was thinking the next three were (2, 23), (2, 29) and (2, 35). My reasoning is the first number is always 2, and the second is (6 *n-1)-1 where n is the set number. However this is incorrect and it does not explain the first set. Can you help me out? Thank you.

 

[tkhunny]

As always, "the next in the series" question is hopelessly conceived. There are infinitely many answers to ANY set of finite beginnings. For this cause, ANYTHING you can justify SHOULD be acceptable as a correct response. Do not spend ANY time trying to find THE answer. It simply doesn't exist.

If you're happy with your response, and can defend it, then go with it. If it gets marked incorrect, fight the grade.

Here's an easy one to justify. (2,f(n)), where n is the number of the ordered pair and f(n) = -(1/3)*[2*n^3 - 18*n^2 + 34*n - 27]. Keep in mind that this is no better than any other sequence you can define.

 

[frappa]

Thank you! Is there a particular method you used to come up with this answer?

 

[HallsofIvy]

Here's one method that gets a completely different answer: label the four pairs "0", "1", "2", "3". Then we are looking for a function, f, such that f(0)= 3, f(1)= 5, f(2)= 11, and f(3)= 17. We can always fit an "n-1" degree polynomial to any n points so we can find a cubic polynomial that gives those four values. Write that polynomial as \(\displaystyle f(x)= ax^3+ bx^2+ cx+ d\).
We must have
f(0)= d= 3
f(1)= a+ b+ c+ d= 5
f(2)= 8a+ 4b+ 2c+ d= 11
f(3)= 27a+ 9b+ 3c+ d= 17.

From the first equation d= 3 so the other 3 are
f(1)= a+ b+ c+ 3= 5 so a+ b+ c= 2
f(2)= 8a+ 4b+ 2c+ 3= 11 so f(2)= 8a+ 4b+ 2c= 8 or 4a+ 2b+ c= 4
f(3)= 27a+ 9b+ 3c+ 3= 17 so f(3)= 27a+ 9b+ 3c= 14

Subtract a+ b+ c= 2 from 4a+ 2b+ c= 4 to get 3a+ b= 2.
Subtract 3 times a+ b+ c= 2 from 27a+ 9b+ 3c= 14 to get 24a+ 6b= 8 or 12a+ 3b= 4.

Subtract 3 times 3a+ b= 2 from 12a+ 3b= 4 to get 3a= -2. a= -2/3.
Then 12a+ 3b= -8+ 3b= 4 so 3b= 12, b= 4.
And a+ b+ c= -2/3+ 4+ c= 10/3+ c= 2 so c= 2- 10/3= -4/3.

f(x)= -(2/3)x^3+ 4x^2- (4/3)x+ 3.

You can check that f(0)= 3, f(1)= -2/3+ 4- 4/3+ 3= 5, f(2)= (-2/3)(8)+ (4)(4)- (4/3)(2)+ 3= -16/3+ 16- 8/3+ 3= 19- 8= 11, and f(3)= (-2/3)(27)+ 4(9)- (4/3)(3)+ 3= -18+ 36- 4+ 3= 39- 22= 17.

And now we can determine f(4)= (-2/3)(64)+ 4(16)- (4/3)(4)+ 3= -128/3+ 64- 16/3+ 3=67- 144/3= 67- 48= 19, f(5 (-2/3)(125)+ 4(25)- (4/3)(5)+ 3= -250/3+ 100- 20/3+ 3= 103- 270/3= 103- 90= 13, and f(6)= (-2/3)(216)+ 4(36)- (4/3)(6)+ 3= -144+ 144- 8+ 3= -5.


So, using this method, the sequence, starting at n= 0, is (2, -(2/3)n^3+ 4n^2- (4/3)n+ 3) and the next three pairs are (2, 19), (2, 13), (2, -5).

 

[ksdhart2]

Yet one more possible sequence that's almost certainly not "the" sequence your instructor had in mind: Define f(1) = 3 and then say that f(n + 1) is the smallest prime number that is both greater than f(n) and not of the form 7 + 6n for any integer n. This choice for f(n) produces:

n	f(n)
1	3
2	5
3	11
4	17
5	23
6	29
7	41

and so on...

 

[Dr.Peterson]

Or look here: http://oeis.org/search?q=3,5,11,17&language=english&go=Search

That lists a few sequences that work, notably the first of each pair of twin primes: 3,5; 5,7; 11,13; 17,19; 29,31; ... .

One other question: you called them "sets", but they are written as ordered pairs. Did you mean that, or something else again?

In any case, the problem is indeed improper. Whoever posed it could just as well have asked, "What have I got in my pocket?"

 

[Steven G]

Can you figure out the next three sets?

(2,3) (2, 5) (2, 11) (2, 17)

I was thinking the next three were (2, 23), (2, 29) and (2, 35). My reasoning is the first number is always 2, and the second is (6 *n-1)-1 where n is the set number. However this is incorrect and it does not explain the first set. Can you help me out? Thank you.

I feel that the 1st finite terms do not have to follow the pattern as long as the rest do follow a pattern. From this view point I feel that your answer is correct. I would certainly challenge my instructor if they marked it wrong!

 

[frappa]

Thank you to everyone who helped with this problem! This problem was actually given by my grandmother and she did actually say any justifiable sequence would count but I found my method to not make sense because it did not account for the first group of numbers (not sure what the difference is between sets and ordered pairs so I'm just going to call them group of numbers haha). Anyway, thank you!