Need help with a number pattern

[Aelfstan]

I can't see the pattern here. It should be easy but it's driving me nuts. Any help?

8 , 12 , 24 , 60 , 168

Thanks

 

[Deleted member 4993]

what pattern are you looking for?

Do you know how to fit a polynomial toa given data set?

 

[Denis]

8 , 12 , 24 , 60 , 168
Did you notice initial differences:
8 (4) 12 (12) 24 (36) 60 (108) 168
4 * 3 = 12 ; 12 * 3 = 36 ....

 

[soroban]

Hello, Aelfstan!

Denis is absolutely correct!


\(\displaystyle \begin{array}{cccccccccc}\text{Sequence} & 8 && 12 && 24 && 60 && 168 \\ \\[-2mm]\text{Differences} && 4 && \underbrace{12}_{3\cdot4} && \underbrace{36}_{3\cdot12} && \underbrace{108}_{3\cdot36} \end{array}\)


$\displaystyle \text{Each difference is 3 times the previous difference.}$

$\displaystyle \text{This can be written: }\;a_n \;=\;a_{n-1} + 3(a_{n-1} - a_{n-2})$

. . $\displaystyle \text{So we have: }\;a_n \;=\;4a_{n-1} - 3a_{n-2}\quad \text{ where }a_1 = 8,\;a_2 = 12$

Each term is 4 times the preceding term minus 3 times the term before that.



$\displaystyle \text{After a }lot\text{ of Recurrence Theory and even more Algebra,}$

. . \(\displaystyle \text{we get: }\:\boxed{a_n \;=\;2\cdot3^{n-1} + 6}\\)


Use this to impress/surprise/terrify your teacher . . .

 

[Denis]

. . \(\displaystyle \text{we get: }\:\boxed{a_n \;=\;2\cdot3^{n-1} + 6}\\)
Use this to impress/surprise/terrify your teacher . . .

Get an extra mark: 2[3^(n-1) + 3] :wink:

 

[Deleted member 4993]

Another expression:

$\displaystyle \frac{4}{3}n^4 \, - \, \frac{32}{3}n^3 \, + \, \frac{104}{3}n^2 \, - \, \frac{136}{3}n \, + \, 28$

But I like Soroban's expression better - it has less constants.

 

[mmm4444bot]

My version ...

$\displaystyle \frac{4}{3}n^4 \, - \, \frac{32}{3}n^3 \, + \, \frac{104}{3}n^2 \, - \, \frac{136}{3}n \, + \, 28$

But I like Soroban's expression better - it has less constants.

Well, I like polynomials a LOT, but (like Soroban) I also wrote a formula to be able to continue the sequence:

8, 12, 24, 60, 168, 492, 1464, ...

I did not post yesterday's result because I have doubts over the original poster's knowledge of recursion; now the cat's out of the bag.

Digging my scrap paper out of the waste-paper basket, I see that I came up with the following.

$\displaystyle a_1 \;=\; 8$

\(\displaystyle a_k \;=\; a_{k-1} + 4 \cdot 3^{k-2} \;\mbox{for k}>1\)

I struggled trying get a formula so that the index k would have a 1-to-1 correspondence with the natural numbers for all elements of the sequence. I gave up after about 45 minutes, settling for k>1.

Therefore, I'm interested with Soroban's final result. Very succinct. (I remember toying with 3^0 + 6, but my factor of 4 kept me from sticking with it becauce the indices broke.)

I'm familiar with the finite-difference method for determining a polynomial that generates all elements of a sequence, but, of course, it does not apply to the sequence in this post as the difference table never reaches a constant row.

(Hmmm -- I wonder how many months it would have taken ME to get from my result to Soroban's ...)

~ Mark