Need help with a number pattern
- Thread starter Aelfstan
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D
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what pattern are you looking for?
Do you know how to fit a polynomial toa given data set?
Do you know how to fit a polynomial toa given data set?
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 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 . . .
D
Deleted member 4993
Guest
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.
\(\displaystyle \frac{4}{3}n^4 \, - \, \frac{32}{3}n^3 \, + \, \frac{104}{3}n^2 \, - \, \frac{136}{3}n \, + \, 28\)
But I like Soroban's expression better - it has less constants.
mmm4444bot
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My version ...
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
Subhotosh Khan said:
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