Number sequence IQ question: What is the next term in this sequence?

[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, ?,

 

[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

 

[Superstring]

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

 

[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.

 

[uart]

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.

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?

 

[uart]

How about the exponential equation,

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

 

[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

 

[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.

 

[alice22]

The next number is 150 apparently.

 

[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

 

[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.

 

[uart]

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

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

 

[uart]

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!

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.

 

[The Chaz]

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.

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

 

[Lord Crc]

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?

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.

 

[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

 

[Dickfore]

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

<br /> x_{n + 3} = 3 x_{n+1} + 2 x_{n} <br />

Indeed:

<br /> 3 \times 8 + 2 \times 3 = 24 + 6 = 30<br />

<br /> 3 \times 18 + 2 \times 8 = 54 + 16 = 70<br />

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

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

 

[The Chaz]

That's reasonable.

 

[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:

<br /> x_{n + p} + c_{1} \, x_{n + p -1} + \ldots + c_{p - 1} \, x_{n + 1} + c_{p} \, x_{n} = 0<br />

has a particular solution of the form:

<br /> x_{n} \propto q^{n}<br />

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

<br /> q^{p} + c_{1} \, q^{p -1} + \ldots + c_{p - 1} \, q + c_{p} = 0<br />

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:

<br /> q^{n} \, P_{s - 1}(n)<br />When we have a complex root:

<br /> q = \rho \, (\cos{\phi} + \textup{i} \, \sin{\phi})<br />

then the particular solution is:

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

The general solution is then:

<br /> 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<br />

The general solution has

<br /> \sum_{\alpha = 1}^{p_{1}} (s_{\alpha} + 1) + \sum_{\beta = 1}^{p_{2}}(2 s_{\beta} + 2) = p + p_{1} + 2 p_{2}<br />

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:

<br /> p = 4, p_{1} = 1, p_{2} = 0 \; \Rightarrow s_{1} = 4<br />

<br /> p = 3, p_{1} = 2, p_{2} = 0 \; \Rightarrow s_{1} + s_{2} = 3 \Rightarrow s_{1} = 2, s_{2} = 1<br />

The first case corresponds to a general solution:

<br /> x_{n} = q^{n} (A \, n^3 + B \, n^{2} + C \, n + D)<br />

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:

<br /> x_{n} = (A n + B) \, q^{n} + C\, p^{n}<br />

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

The characteristic polynomial in this case is:

<br /> \begin{array}{l}<br /> (q + 1)^{2} \, (q - 2) = 0 \\<br /> <br /> q^{3} - 3 q - 2 = 0<br /> \end{array}<br />

corresponding to the recurrence relation stated in my previous post.

 

[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/

 

[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!

 

[uart]

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

A linear homogeneous recursion of order p:

<br /> x_{n + p} + c_{1} \, x_{n + p -1} + \ldots + c_{p - 1} \, x_{n + 1} + c_{p} \, x_{n} = 0<br />

has a particular solution of the form: ...

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

First let's 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 let's 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}

 

[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).

 

[Dickfore]

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

<br /> x_{k+1} = a_{0} + a_{1} \, x_{k} + a_{2} \, x_{k}^{2} + a_{3} \, x_{k}^{3}<br />

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

with the solution:

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

which leads to the solution

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

which would lead to the next element being:
<br /> x_{6} = \frac{5002618}{4455}<br />

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

 

[sjb-2812]

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!

Pizza slicing, by any chance?

 

[shinkyo00]

Here's my solution.

Sequence: 5, 8, 18, 30, 70, ?

Answer: 118

Explanation:

5 * 2 - 2 = 8, apply to (-2): + (-2 * -2) <-- +4

8 * 2 + 2 = 18, apply to (+2): + (4 * -2) <-- -8

18 * 2 - 6 = 30, apply to (-6): + (-8 * -2) <-- +16

30 * 2 + 10 = 70, apply to (+10): + (16 * -2) <-- -32

70 * 2 - 22 = 118

So there, try 118 and see if it works. It should. This is the simplest answer yet at the same time is most complete. Notice that +4, -8, +16 is the real pattern here, and it is repeated exactly once (i.e. +4 * -2 = -8 is the first case, then -8 * -2 = +16 is the repetition).

Now, which test did you get this question from?

EDIT:

I overlooked the OP's reply that 150 is the correct answer. I guess that invalidates mine!

 

[Dosmascerveza]

I just looked at it for a while and just when i was about to give up and cheat i thought let's look at this thing as 3 pieces and then boom 150 popped into my head. I fail at math.

 

[Phenomenal]

This is being WAY overanalysed in my opinion. In terms of this being an IQ question the logical answer I can understand being 150.

3 x 2 + 2 = 8
8 x 2 + 2 = 18
18 x 2 - 6 = 30, does not follow rule.
30 x 2 + 10 = 70

70 x 2 + 10 = 150

Here we have to make the assumption that it's going to be +10 again to follow the previous. If anyone wants to reason on why we wouldn't assume that I'd welcome it. I see this as being the only logical way of looking at the question.

 

[uart]

Here's my solution.
Sequence: 5, 8, 18, 30, 70, ?

Not good. You've given a solution to a different problem, or changed the first term in the sequence to make it fit your solution. The first term was supposed to be 3.

 

[uart]

This is being WAY overanalysed in my opinion. In terms of this being an IQ question the logical answer I can understand being 150.

3 x 2 + 2 = 8
8 x 2 + 2 = 18
18 x 2 - 6 = 30, does not follow rule.
30 x 2 + 10 = 70

70 x 2 + 10 = 150

Here we have to make the assumption that it's going to be +10 again to follow the previous. If anyone wants to reason on why we wouldn't assume that I'd welcome it. I see this as being the only logical way of looking at the question.

So what would the next term after 150 be Phenomenal. Would it follow the "rule" or not follow the rule? It's not such a good rule if it's only sometimes followed.

 

[Phenomenal]

So what would the next term after 150 be Phenomenal. Would it follow the "rule" or not follow the rule? It's not such a good rule if it's only sometimes followed.

The point is you're only asked for the NEXT number. It doesn't ask you to work out the next several.. Excuse my use of the term rule, you're correct. What I am implying is that this is the only 'pattern' that really applies and thus 150 makes perfect sense.

This is one question in an IQ test as already said, it's not expected that the situation is overanalysed using the previous calculations that have already been done so far in this thread.

 

[The Chaz]

The point is you're only asked for the NEXT number. It doesn't ask you to work out the next several.. Excuse my use of the term rule, you're correct. What I am implying is that this is the only 'pattern' that really applies and thus 150 makes perfect sense.

This is one question in an IQ test as already said, it's not expected that the situation is overanalysed using the previous calculations that have already been done so far in this thread.

Uart has failed to justify these extensive calculations in the context of an IQ test and - more generally - hasn't addressed any of my three points from 9:46am a few weeks back. (can't see the date or the post number on my iPhone)...
I'll let you omit your response to my #1 ;)

 

[Dickfore]

I can only find this justification for 150. If we write down the differences:

<br /> \begin{array}{cc}<br /> 3 &amp; \\<br /> &amp; 5 \\<br /> 8 &amp; \\ <br /> &amp; 10 \\<br /> 18 &amp; \\<br /> &amp; \mathbf{12} \\<br /> 30 &amp; \\<br /> &amp; 40 \\<br /> 70 &amp; \\<br /> \end{array}<br />

I see that the twelve "sticks out" from the sequence. It would be much nicer if it were a twenty. And, indeed, in many languages twelve and twenty are similar words and small children often make the mistake of confusing them (in English, a similar situation arises from the -teen's and the -ty's).

If it were 20 instead of 12, then the first differences would form a geometric progression with a quotient 2 and the next term in this sequence (the first differences) would be 80.

But, the next term in the original sequence is equal to the sum of the previous term and the last first difference, namely:

70 + 80 = 150

 

[The Chaz]

As to #3 from a previous post of mine...
1,2,4,8,16,31 is justified as the number of pieces in a circle, when segments are drawn between 1,2,3,4,5, and 6 points on its circumference.

"Number Freak" by Niederman gives explanations for how "30" and "33" could also complete the sequence
1,2,4,8,16__

30: the number of divisors in (6!).
33: the number of ways in which the first player gets killed in a five-player Russian Roulette game using a gun having n chambers, where the number of bullets can equal anything from 1 to n, with no rotations of the cylinder allowed.

 

[Chris11]

LOL. The pattern detecting system couldn't even detect the following pattern: 2, 4, 8, 16,..., 2^n

 

[Noxide]

3, 8, 18 , 30, 70, 150 , 300, 690, 1470, ...dumb question

 

[Dickfore]

3, 8, 18 , 30, 70, 150 , 300, 690, 1470, ...


dumb question

would you mind elaborating how you obtained this continuation of the sequence?

 

[Noxide]

would you mind elaborating how you obtained this continuation of the sequence?

A = 3, 8, 18, 30, 70, 150, 300, ... <=> A = a, b, d, e, f, h, i, ...

a = 3 (arbitrary)
b = 8 (arbitrary)
b-a = c
b + 2c = d
e = 10a
f = 10(b-1)
f-e = g
2g + f = h
10e = i
...

i.e. the sequence is uniquely determined by a given a and b such that sequence a, b, c, ... = A

A = a, b, b+2(b-a), 10a, 10(b-1), 2[10(b-1) - 10a] + 10(b-1), 10(10a), 10[10(b-1)], 2{10[10(b-1)]- 10(10)a}, ...

 

[Dickfore]

A = 3, 8, 18, 30, 70, 150, 300, ... <=> A = a, b, d, e, f, h, i, ...

a = 3 (arbitrary)
b = 8 (arbitrary)
b-a = c
b + 2c = d
e = 10a
f = 10(b-1)
f-e = g
2g + f = h
10e = i
...

i.e. the sequence is uniquely determined by a given a and b such that sequence a, b, c, ... = A

A = a, b, b+2(b-a), 10a, 10(b-1), 2[10(b-1) - 10a] + 10(b-1), 10(10a), 10[10(b-1)], 2{10[10(b-1)]- 10(10)a}, ...

You can combine your definition of c and d, to eliminate the intermediate value c (and g further on):
<br /> d = b + 2c = b + 2(b - a) = 3b - 2a<br />
Other than that, you logic seems like stabbing in the dark to find the given answer 150.

 

[Chris11]

Moral of the Story: IQ tests are meaningless

 

[Noxide]

You can combine your definition of c and d, to eliminate the intermediate value c (and g further on):
<br /> d = b + 2c = b + 2(b - a) = 3b - 2a<br />
Other than that, you logic seems like stabbing in the dark to find the given answer 150.

Of course I can, and I did: A = a, b, b+2(b-a), 10a, 10(b-1), 2[10(b-1) - 10a] + 10(b-1), 10(10a), 10[10(b-1)], 2{10[10(b-1)]- 10(10)a}, ...
A is defined purely in terms of a and b

How is it stabbing in the dark?
The difference between the first two terms is half the difference between the second and third terms. The sequence has units that are triplets...

 

[Dickfore]

Of course I can, and I did: A = a, b, b+2(b-a), 10a, 10(b-1), 2[10(b-1) - 10a] + 10(b-1), 10(10a), 10[10(b-1)], 2{10[10(b-1)]- 10(10)a}, ...
A is defined purely in terms of a and b

How is it stabbing in the dark?
The difference between the first two terms is half the difference between the second and third terms. The sequence has units that are triplets...

What's the logic for 10(b - 1) in the second term of the second triplet, i.e. 70? If you followed any logic from the previous one, you should have written 80.

 

[The Chaz]

What's the logic for 10(b - 1) in the second term of the second triplet, i.e. 70? If you followed any logic from the previous one, you should have written 80.

Maybe this is less a dumb question than a dumb answer...?
That this comes from an IQ test is the only justification for considering the (or a) pattern observed in the first three terms and applying that to the next set of three.

This reminds me of the analogy questions on the old SAT. Something like
fish:water::telephone:_______

 

[Dickfore]

Let us use polynomial inetrpolation:

<br /> \begin{array}{ccccc}<br /> 3&amp; &amp; &amp; &amp; \\<br /> &amp;5&amp; &amp; &amp; \\<br /> 8&amp; &amp;5&amp; &amp; \\<br /> &amp;10&amp; &amp;-3 \\<br /> 18&amp; &amp;2&amp; &amp;19\\<br /> &amp;12&amp; &amp;16&amp; \\<br /> 30&amp; &amp;18&amp; &amp;\mathbf{19}\\<br /> &amp;40&amp; &amp;35&amp; \\<br /> 70&amp; &amp;53&amp; &amp; \\<br /> &amp;93&amp; &amp; &amp; \\<br /> \mathbf{163}&amp; &amp; &amp; &amp; <br /> \end{array}<br />

 

[uart]

Of course I can, and I did: A = a, b, b+2(b-a), 10a, 10(b-1), 2[10(b-1) - 10a] + 10(b-1), 10(10a), 10[10(b-1)], 2{10[10(b-1)]- 10(10)a}, ...
A is defined purely in terms of a and b

How is it stabbing in the dark?
The difference between the first two terms is half the difference between the second and third terms. The sequence has units that are triplets...

That's ok, but the following "triplet" solution is no more (or no less) arbitrary.

3 8 18 : 30 70 2097 : 57 132 7506 :

The third term in each triplet is the product of the previous two terms minus the digit sum (of that product). For example 3*8=24 and 24-2-4 = 18.

The triplets are continued via T_{3n+1} = T_{3n-2} + 27 and T_{3n+2} = T_{3n-1}+62, starting with T1 = 3 and T2 = 8.

 

[Tedjn]

I've got an even better idea. Let's plot the sequence against n, draw a random curve through those points, and then look at the graph! An experimental way to decide it once and for all.

 

[Phenomenal]

Maybe this is less a dumb question than a dumb answer...?
That this comes from an IQ test is the only justification for considering the (or a) pattern observed in the first three terms and applying that to the next set of three.

This reminds me of the analogy questions on the old SAT. Something like
fish:water::telephone:_______

Funny how the logical explanation has yet again been ignored and overlooked by you guys.

I assume you're just trying to worth out a real "pattern" in these numbers now to prove there IS indeed a pattern, correct?

Without logic we are adrift on a sea of rationalizations.

 

[Count Iblis]

http://www.mat.univie.ac.at/~kratt/rate/rate.html

 

[Noxide]

That's ok, but the following "triplet" solution is no more (or no less) arbitrary.

3 8 18 : 30 70 2097 : 57 132 7506 :

The third term in each triplet is the product of the previous two terms minus the digit sum (of that product). For example 3*8=24 and 24-2-4 = 18.

The triplets are continued via T_{3n+1} = T_{3n-2} + 27 and T_{3n+2} = T_{3n-1}+62, starting with T1 = 3 and T2 = 8.

This cannot be the case... You are overlooking an axiom of the sequence. That axiom being the difference between the first two terms of the triplet is half of the difference between the 2nd and third terms of each triplet.

 

[Dickfore]

This cannot be the case... You are overlooking an axiom of the sequence. That axiom being the difference between the first two terms of the triplet is half of the difference between the 2nd and third terms of each triplet.

lolwut? Where was this axiom postulated?