# Number sequence IQ question.

1. May 23, 2010

### alice22

This sequence cropped up in an IQ test.

I can't for the life of me work out the next term, any ideas?

3, 8, 18, 30, 70, ?,

2. May 23, 2010

### Mute

You're sure that's the sequence? The online encyclopedia of integer sequences turns up nothing. Even dropping the 70, only one sequence turns up, and the 3 is the second number in that sequence (and that sequence doesn't seem to have a closed form expression).

http://www.research.att.com/~njas/sequences/index.html [Broken]

Last edited by a moderator: May 4, 2017
3. May 23, 2010

### Superstring

This says 150: http://www.patternsolver.com/ [Broken] but I was trying to figure it out for well over 20 minutes with no success.

Last edited by a moderator: May 4, 2017
4. May 23, 2010

### Lord Crc

The difference between the second and the third number is twice that of the difference between the second and the first. Thus 70-30 = 40, and 2*40+70 = 150.

5. May 24, 2010

### uart

However 30 is NOT equal to 18 + 2 x 10

And 70 is NOT equal to 30 + 2 x 12

I'm not sure how you managed to overlook that?

Last edited: May 24, 2010
6. May 24, 2010

### uart

$$Q(n) = \frac{1}{3} \left( 7 \times 1^n - 33 \times 2^n + 66 \times 3^n - 40 \times 4^n + 9 \times 5^n \right)$$

You can verify that Q(0) = 3, Q(1) = 8, Q(2) = 18, Q(3) = 40 and Q(4) = 70.

So we could fill in the next term with Q(5) = 718

7. May 24, 2010

### uart

Or if you prefer polynomials then how about,

$$P(n) = \frac{1}{24} \left( 840 - 1642 n + 1147 n^2 - 302 n^3 + 29 n^4 \right)$$

You can verify that P(1) = 3, P(2) = 8, P(3) = 18, P(4) = 40 and P(5) = 70.

So we could fill in the next term with P(6) = 193

Last edited: May 24, 2010
8. May 24, 2010

### Phrak

3, 8, 18, 30, 70,

This obviously the solution for the zeros of the polynomial equation y=(x-3)(x-8)(x-18)(x-30)(x-70)(x-29pi)=0 in rank sequence. So the next number is 29*pi and that ends the series.

9. May 24, 2010

### alice22

The next number is 150 apparently.

10. May 24, 2010

### Xitami

$R(n)=\frac{1}{15}\left(-45(-1)^n+36(-2)^n+63(2)^n-(3)^n-8(-3)^n\right)$
R(5)=174

11. May 24, 2010

### The Chaz

You all seem to have missed that this is from an IQ TEST.
Yes, you can write an infinite list of polynomials with these roots, but that's not the question.
FIND.
THE.
PATTERN!!!!!!!!!!!!!

I'll give you one whose solution I happen to know.
3, 4, 3, 5, 3, 6, 8, 7, ....
Yes, I am sure about the "8". This isn't about numerical patterns per sé.

12. May 24, 2010

### uart

Hi Xitami. R(0) through to R(3) work but R(4)=54? It should be 70 so that exponential equation doesn't work.

13. May 24, 2010

### uart

The functions Q(x) and P(x) that I posted are number patterns Chaz. They might not be the pattern that you want, but they most definitely are number patterns in the usual sense.

14. May 24, 2010

### The Chaz

1) It's The Chaz...
2) Nobody is arguing that Q and P aren't "number patterns". But you've gotten so skilled at solving problems that you have solved a problem of your own imagination. The relevant problem is found in an IQ test, which leads us to preclude certain "solutions" (including, and especially, those like you suggest).
3) Why don't you give (3,4,3,5,3,6,8,7....) a shot? It is VERY simple, but difficult to spot. Maybe these kind of questions aren't fit for a "math" forum. Or maybe "math" needs to learn to think outside the box a little. Click and drag for hints...

a) linguistic

b) mod2

15. May 24, 2010

### Lord Crc

I didn't overlook that. I assumed that the pattern didn't have to relate each number in the sequence to every other number. In this case I believe the pattern relates each triple individually, ie the numbers in the first triple is independent of the previous one. However the equation used with each triple is not.

As usual with these sequence questions in IQ tests, there's an assumption that the numbers have some simple "logical" relation, and aren't just roots of an arbitrary polynomial or similar, as then the next number could be anything.

16. May 24, 2010

### Xitami

$R(n)=\frac{1}{15}\left(-35(-1)^n+24(-2)^n+59(2)^n+(3)^n-4(-3)^n\right)$

$R(5)=158$

17. May 24, 2010

### Dickfore

I have found that the sequence of these 5 numbers satisfies the following recursion:

$$x_{n + 3} = 3 x_{n+1} + 2 x_{n}$$

Indeed:

$$3 \times 8 + 2 \times 3 = 24 + 6 = 30$$

$$3 \times 18 + 2 \times 8 = 54 + 16 = 70$$

We can conclude that the next number is:
$$x = 3 \times 30 + 2 \times 18 = 90 + 36 = 126$$

So, my guess is that the next nymber in the sequence is 126.

18. May 24, 2010

### The Chaz

That's reasonable.

19. May 24, 2010

### Dickfore

To justify my solution with more than just "looking at the stars", I will offer this reasoning.

A linear homogeneous recursion of order p:

$$x_{n + p} + c_{1} \, x_{n + p -1} + \ldots + c_{p - 1} \, x_{n + 1} + c_{p} \, x_{n} = 0$$

has a particular solution of the form:

$$x_{n} \propto q^{n}$$

for those values of q that are roots to the characteristic equation:

$$q^{p} + c_{1} \, q^{p -1} + \ldots + c_{p - 1} \, q + c_{p} = 0$$

This polynomial has exactly p complex roots (the fundamental theorem of algebra). However, because the coefficients of the equation are real, it follows that if $q$ is a root, then so is $q^{\ast}$. This means that the roots are either real or come in complex conjugate pairs. When we have a root of multiplicity s, the particular soluiton corresponding to it is:

$$q^{n} \, P_{s - 1}(n)$$

When we have a complex root:

$$q = \rho \, (\cos{\phi} + \textup{i} \, \sin{\phi})$$

then the particular solution is:

$$\rho^{n} \, \left( A_{s - 1} (n) \, \cos(n \phi) + B_{s - 1}(n) \, \sin(n \phi) \right)$$

The general solution is then:

$$x_{n} = \sum_{\alpha = 1}^{p_{1}} {q_{\alpha}^{n} \, P_{s_{\alpha} - 1}(n) } + \sum_{\beta = 1}^{p_{2}} {\rho_{\beta}^{n} \, \left( A_{s_{\beta} - 1}(n) \, \cos(n \phi_{\beta}) + B_{s_{\beta} - 1}(n) \, \sin(n \phi_{\beta}) \right) } , \: \sum_{\alpha = 1} ^{p_{1}} {s_{\alpha}} + 2 \sum_{\beta = 1}^{p_{2}} {s_{\beta}} = p$$

The general solution has

$$\sum_{\alpha = 1}^{p_{1}} (s_{\alpha} + 1) + \sum_{\beta = 1}^{p_{2}}(2 s_{\beta} + 2) = p + p_{1} + 2 p_{2}$$

parameters. Since we have 5 points of the series, we can solve for the coefficients if there are 5 unknowns. How can we have 5 unknowns? These are the only possibilities:

$$p = 4, p_{1} = 1, p_{2} = 0 \; \Rightarrow s_{1} = 4$$

$$p = 3, p_{1} = 2, p_{2} = 0 \; \Rightarrow s_{1} + s_{2} = 3 \Rightarrow s_{1} = 2, s_{2} = 1$$

The first case corresponds to a general solution:

$$x_{n} = q^{n} (A \, n^3 + B \, n^{2} + C \, n + D)$$

Using the conditions and Mathematica, it turns out there is no solution with rational values for A, B, C, D and q.

The other case corresponds to a general solution:

$$x_{n} = (A n + B) \, q^{n} + C\, p^{n}$$

It turns out that there is a solution with rational values in this case:
$$A = \frac{4}{3}, \ B = -\frac{10}{9}, \ q = -1,\ C = \frac{37}{9}, \ p = 2$$

The characteristic polynomial in this case is:

$$\begin{array}{l} (q + 1)^{2} \, (q - 2) = 0 \\ q^{3} - 3 q - 2 = 0 \end{array}$$

corresponding to the recurrence relation stated in my previous post.

20. May 24, 2010

### Jerbearrrrrr

Is there a good reason as to why it's 150?

Raises some good questions about patterns though. Is the only reason that the solution to 4,5,6,7,_ is 8, is that 8 is one that anyone can understand? Is the kid that remarks upon the infinity of solutions to a badly defined problem smarter than the one who puts 8?
How valid a concept is IQ anyway?

Last time I did an IQ test I came out with 100, which means I'm exactly average and therefore don't really have anything to live up to \o/

Last edited: May 24, 2010
21. May 24, 2010

### The Chaz

You make a good point. There is a subjective element to this, which is where the hardline polynomial advocates and I part ways. I once saw a justification for 31 as the next in the sequence:
1,2,4,8,16,...
And it made sense, but you'd better not put that on the IQ test!!!

22. May 25, 2010

### uart

Good solution dickfore. You can also get that result from simple linear algebra.

First lets check if there are any recurrence solutions length 2. (using matlab syntax)

1. Set up the matrix M = [3,8,18; 8,18,30; 18,30,70];

2. Row reduce. R = rref(M). Which in this case returns R = [1 0 0 ; 0 1 0 ; 0 0 1].

This is three equations in two unknowns, so unless R has a full zero row there will be no solutions. In this case R is the identity matrix so there is no solution.

Now lets check for length three solutions

1. Set up the matrix M = [3,8,18,30; 8,18,30,70];

2. Row reduce. R = rref(M). Which in this case returns R = [1 0 -8.4 2; 0 1 5.4 3].

This time we have two eqns in three unknowns, so there will either be zero solutions or an infinite number of solutions. In this particular case it does have solutions as follows,

$$(a_0, a_1, a_2) = (2 + \frac{42t}{5}, \, 3 - \frac{27t}{5}, \, t)$$

So there are an infinite number of length three recurrence relations (one for each value of "t") of the form :

$$x_{k+3} = a_0 \, x_k + a_1 \, x_{k+1} + a_2 \, x_{k+2}$$

Last edited: May 25, 2010
23. May 25, 2010

### uart

BTW if anyone is interested in how I obtained the exponential and polynomial solutions, Q(n) and P(n), in replies #5 and #6 it is as follows. Again it's linear algebra (using matlab syntax).

1. Set up the row vector x = [1, 2, 3, 4, 5]

2. Set up the matrix M = [x.^0; x.^1; x.^2; x.^3; x.^4]

3. Set up the colum vector y = [3; 8; 18; 30; 70]

3. The coefficients of the exponential equation Q(n) are the solutions to the linear equation M t = y. That is, t = inv(M) y.

4. The coefficients of the polynomial equation P(n) are the solutions to the linear equation M' t = y. That is, t = inv(M') y. (Where M' is the transpose of M).

24. May 25, 2010

### Dickfore

On the other hand, we might have a first order, but nonlinear relation:

$$x_{k+1} = a_{0} + a_{1} \, x_{k} + a_{2} \, x_{k}^{2} + a_{3} \, x_{k}^{3}$$

which gives the system of eqations:
$$\left[ \begin{array}{cccc} 1 & 3 & 3^{2} & 3^{3} \\ 1 & 8 & 8^{2} & 8^{3} \\ 1 & 18 & 18^{2} & 18^{3} \\ 1 & 30 & 30^{2} & 30^{3} \end{array}\right] \cdod \left[ \begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \\ a_{3} \end{array} \right] = \left[ \begin{array}{c} 8 \\ 18 \\ 30 \\ 70 \end{array} \right]$$

with the solution:

$$(a_{0}, a_{1}, a_{2}, a_{3}) = (-\frac{278}{165}, \frac{28382}{7425}, -\frac{4784}{22275}, \frac{124}{22275})$$

$$x_{k + 1} = \frac{2\, (62 \, x_{k}^{3} - 2392 \, x_{k}^{2} + 42573 \, x_{k} - 18765)}{22275}$$

which would lead to the next element being:
$$x_{6} = \frac{5002618}{4455}$$

But, this is not an integer solution, so we would suspect this is the way to go.

25. May 25, 2010

### sjb-2812

Pizza slicing, by any chance?