I need help solving this number pattern:?

Question

1, 2, 8, 22, 47, ?, ?, ?

Answer 1

97

Answer 2

Could it be a quadratic pattern? Let's conjecture this.

Let's come up with a sequence, f(n), where
f(n) = an^2 + bn + c
f(1) = 1 {i.e. the first term is 1}
f(2) = 2 {i.e. the second term is 2}
f(3) = 8 {i.e. the third term is 8}

Now, we state f(1), f(2), and f(8) by their definitions, and equate it to what we want them to be equal to.

f(1) = a + b + c = 1
f(2) = 4a + 2b + c = 2
f(3) = 9a + 3b + c = 8

System of equations; three equations three unknowns. Without showing you the details of what I did to solve it, I get the following solution: a = 5/2, b = -13/2, c = 5

Therefore, our function is
f(n) = (5/2)n^2 - (13/2)n + 5

Let's test it out.
f(1) = (5/2) - (13/2) + 5 = -8/2 + 5 = -4 + 5 = 1
f(2) = (5/2)(4) - (13/2)(2) + 5 = 10 - 13 + 5 = 2
f(3) = (5/2)(9) - (13/2)(3) + 5 = 45/2 - 39/2 + 5 = 6/2 + 5 = 8

Now, for the moment of truth... we WANT f(4) to equal 47.

f(4) = (5/2)(16) - (13/2)(4) + 5 = 40 - 26 + 5 = 45 - 26 = 19

Guess not. :( But at least I tried.

Edit: Your question has me obsessing. How about an exponential relationship? That is, maybe

f(n) = b * a^n + c.

Therefore,
f(1) = ab + c = 1
f(2) = (a^2)b + c = 2
f(3) = (a^3)b + c = 8

Three equations, three unknowns. Solution (details omitted) is
a = 6, b = 1/30, c = 4/5

f(n) = (1/30) * 6^n + 4/5.
This is actually reducible, to
f(n) = (1/5)(1/6)(6^n) + 4/5
f(n) = (1/5)(6^(n-1)) + 4/5

Let's try.
f(1) = (1/5)(1) + 4/5 = 5/5 = 1
f(2) = (1/5)(6^1) + 4/5 = 6/5 + 4/5 = 10/5 = 2
f(3) = (1/5)(6^2) + 4/5 = (1/5)(36) + 4/5 = 40/5 = 8
Moment of truth,
f(4) = (1/5)(6^3) + 4/5 = 216/5 + 4/5 = 220/5 = 44

So friggin' close, but no! There isn't an exponential relationship, either!

I don't feel like conjecturing a cubic or quartic relationship; not in any mood to solve for 4-equation, 4-unknowns.

Answer 3

Take a look at the amount of the increase to get from each term of the sequence to the next.

These numbers form the sequence 1,6,14 and 25.

Now do the same to this new sequence and you get 5, 8, 11. There's a pattern here of 3 being added each time. So, continuing the pattern we get 5, 8, 11, 14, 17, 20.

Working backwards to the second sequence we get 1, 6, 14, 25, 39, 56, 76.

And finally using these values in the original sequence gives:

1, 2, 8, 22, 47, 86, 142, 218

Answer 4

The answer that I got was: 86. This is how I got it:

1 2 8 22 47 86 your pattern
\ /\ /\ /\ /\ /
1 6 14 25 39 if you find the differences
\ /\ /\ /\ /
5 8 11 14 every third number

Hope this helps...don't know if it is right...but it took me 20 minutes to figure it out!!

It is a mobius pattern it has to repeat or reverse it's self in order to continue the 3 pattern on the tertiary line.

Answer 5

The difference pattern of 1, 2, 8, 22, 47,(86, 142, ...?) ..., is
1, 6, 14, 25,(39, 56, ...?)...

The difference pattern of 1, 6, 14, 25,..., is,
5, 8, 11, (14, 17, 20...?)

1, 2, 8, 22, 47, 86, 142, ...

It sounds not unique.

Answer 6

n = 2 - x - 0.5x^2 + 0.5x^3
1, 1
2, 2
3, 8
4, 22
5, 47
6, 88
7, 142
8, 218