Find the next three sequences

Question

Please find the next three numbers in the sequence
The sequence is ...
3 , 4 , 7 , 11 , 18 , 27 , __ , __ , __.

please look at the sequence carefully, and I would also appreciate if you could also give me the rule of how you found it.

I am sure the number is 27 not 29 so please do not give me a sequence with 29

Answer 1

Hi Bon--

 

I'm with Mike in that the original question really HAD to be 29, not 27, but can you tell us what grade or level of math you're doing, and/or give us other examples of problems you've been working on?

 

Also, can you tell us where the problem came from (i.e., a textbook or website, or did a teacher make it up?)

 

I'd be pretty surprised if you were expected to solve it in any of the ways Mike laid out (not that I have any idea what any of that meant, Mike!), but I suppose anything is possible.

 

Good luck.

 

Matt up in Westchester

Answer 2

Hi Bon,

 

29 would be absolutely great.  There are many answers to your question, none of them obvious.  Here's one: 

List the primes in order, so that p(1) = 2, p(2) = 3, p(3) = 5, p(4) = 7, ....

 

Then the sequence is given by s(0) = 3, s(n) = s(0) + sum(j=1...n)(1 + (p(j+18)-p(19)) / 2). 

 

Here's some more detail on that one:

 

p(19), p(20), .... = {67, 71, 73, 79, 83, 89, 97, 101,...}

 

sum(j=1,1) = 1 + (67-67)/2 =1

sum(j=1,2) = 1 + 1+(71-67)/2   = 1 + 3 = 4

sum(j = 1,2,3) = 4 + 1 + (73-67)/2 = 4 + 4 = 8

sum(j = 1,2,..4) =  8 + 1 + (79-67)/2 = 8 + 7 = 15

sum(j = 1,2,..5) = 15 + 1 + (83-67)/2 = 15 + 9 = 24

sum(j = 1,2,..6) = 24 + 1 + (89-67)/2 = 24 + 12 = 36

sum(j = 1,2,..7) =  36 + 1 + (97-67)/2 = 36 + 16 = 52

sum(j = 1,2,..8) =  52 + 1 - (101-67)/2 = 52 + 18 = 70

 

So sum(j=1,...)(1 + (p(j+18)-p(19)) / 2) = (1, 4, 8, 15, 24, 36, 52, 70...)

s(0) = 3, 3 + sum(j=1,...) = (4, 7, 11, 18, 27, 39, 55, 73...)

 

So the next three terms would be 39, 55, 73.

 

Another possibility is phi(n), Euler's totient function.  It is the count of positive integers <= n that are relatively prime to n.  For example, phi(4) = 2, since 4 and 3 are relatively prime, as are 4 and 1, but 4 and 2 share common factors as do 4 and 4.  phi(4,5,6,7,8,9,10,11...) = 4,2,6,4,6,4,10...

 

In that case, the formula for the sequence is given by s(0) = 3, s(n) = s(0) + sum(j=1...n)(1 + (phi(j+4)-phi(5)) / 2).

The next three terms of that sequence are 39, 53, 72.

 

Yet another way is to define a polynomial that takes the value f(1) = 3, f(2) = 4, f(3) = 7,...

That turns out to be 14 - x*71.35 / 3.0 + x^2 * 17.625 - x^3 * 17.0/3.0 + 0.875 x^4 - 0.05*x^5

In that case, the next three terms are 28, -4, -115.

 

But for my money, the person who gave you the sequence should have written down 27.

 

Hope that helps.

 

Regards,

Mike N.