Need help figuring out this sequence...

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October 8th 2007, 09:40 AM
snakeyster

Need help figuring out this sequence...

Basicaly, I need help on this sequence, as my homework is due in tomorrow. One question on sequences is:

1
3
11
67

I'm supposed to find out what comes next. Can anyone help? Even if you tell me how you got the answer, that would be a great help
October 8th 2007, 10:00 AM
CaptainBlack

Quote:

Originally Posted by snakeyster

Basicaly, I need help on this sequence, as my homework is due in tomorrow. One question on sequences is:

1
3
11
67

I'm supposed to find out what comes next. Can anyone help? Even if you tell me how you got the answer, that would be a great help

Since the online encyclopedia of integer sequences knows this not, I would
suggest that this is not a mathematicaly defined sequence but is somehow
connected with the site you got it from

RonL
October 8th 2007, 10:33 AM
Jhevon

i think $a_n = 7n^3 - 39n^2 + 70n - 37$ for $n \in \mathbb {N}$ works

thus, the fifth term is: 213

(i found the cubic that hit the points (1,1), (2,3), (3,11) and (4,67) by solving a system of equations)
October 8th 2007, 10:42 AM
Soroban

Hello, snakeyster!

Quote:

I need help on this sequence: . $1,\,3,\,11,\,67,\,\cdots$

I see a pattern . . .

. . $\begin{array}{ccccc} a_1 & = & 1 & & \\ a_2 & = & {\color{red}2}(a_1)\,{\color{red}+\,1} & = & 3 \\ a_3 & = & {\color{red}4}(a_2)\,{\color{red}-\,1} & = & 11 \\ a_4 & = & {\color{red}6}(a_3)\,{\color{red}+\,1} & = & 67\end{array}$

I bet the next one is:

. . $\begin{array}{ccccc} a_5 & = & {\color{red}8}(a_4)\,{\color{red}-\,1} & = & 535\end{array}$

The general term seems to be: . $a_n\;=\;2(n-1)(a_{n\text{-}1}) + (\text{-}1)^n$ . for $n \geq 2$
October 8th 2007, 10:43 AM
Jhevon

Quote:

Originally Posted by Soroban

Hello, snakeyster!

I see a pattern . . .

. . $\begin{array}{ccccc} a_1 & = & 1 & & \\ a_2 & = & {\color{red}2}(a_1)\,{\color{red}+\,1} & = & 3 \\ a_3 & = & {\color{red}4}(a_2)\,{\color{red}-\,1} & = & 11 \\ a_4 & = & {\color{red}6}(a_3)\,{\color{red}+\,1} & = & 67\end{array}$

I bet the next one is:

. . $\begin{array}{ccccc} a_5 & = & {\color{red}8}(a_4)\,{\color{red}-\,1} & = & 535\end{array}$

The general term seems to be: . $a_n\;=\;2(n-1)(a_{n\text{-}1}) + (\text{-}1)^n$ . for $n \geq 2$

hmm, we got different answers. i guess i made a mistake then. oh well, good job! I don't know how you see these pattens
October 8th 2007, 01:23 PM
CaptainBlack

Quote:

Originally Posted by Jhevon

hmm, we got different answers. i guess i made a mistake then. oh well, good job! I don't know how you see these pattens

You made no mistake, one can chose the next term to have an arbitary
value and find a polynomial that will take all the given values and our arbitary
next value.

Your cubic does give the first four terms, and so is as valid a rule for the
next term as any other, and I suspect it involves no more arbitary constants
than does Soroban's

RonL
October 8th 2007, 01:23 PM
topsquark

Quote:

Originally Posted by Jhevon

hmm, we got different answers. i guess i made a mistake then. oh well, good job! I don't know how you see these pattens

You can find an infinite number of polynomial fits to a finite data set. So the fact that you and Soroban came up with different answers is not a surprise.

But I'm with you. I'd love to know how he comes up with this stuff! :)

-Dan
October 8th 2007, 01:24 PM
Jhevon

Quote:

Originally Posted by CaptainBlack

You made no mistake, one can chose the next term to have an arbitary
value and find a polynomial that will take all the given values and our arbitary
next value.

Your cubic does give the first four terms, and so is as valid a rule for the
next term as any other, and I suspect it involves no more arbitary constants
than does Soroban's

RonL

Quote:

Originally Posted by topsquark

You can find an infinite number of polynomial fits to a finite data set. So the fact that you and Soroban came up with different answers is not a surprise.

But I'm with you. I'd love to know how he comes up with this stuff! :)

-Dan

Pfft! sequences are so weird!
October 8th 2007, 01:52 PM
Soroban

HellO!

RonL and Dan are, of course, absolutely correct.

A sequence of numbers is meaningless unless there is a stipulation,
. . a promise that that the sequence "continues in a similar manner",
. . that there is indeed a "reasonable" pattern to the terms.

I've posted the following before, but I enjoy showing it off.

Find the next term of the sequence: . $\bf{1,\,3,\,5,\,7,\,\cdots}$

Answer: 8

Justification
I was using this function: . $f(n) \;=\;-\frac{1}{24}\left(n^4 - 10n^3 + 35n^2 - 98x + 48\right)$

"That's unfair!" you say?
Okay, here's a better explanation . . .

The sequence: . $1,\,3,\,5,\,7,\,8,\,\cdots$ is the sequence of natural numbers
. . whose English names contain the letter "e".
October 8th 2007, 11:02 PM
CaptainBlack

Quote:

Originally Posted by Soroban

HellO!

RonL and Dan are, of course, absolutely correct.

A sequence of numbers is meaningless unless there is a stipulation,
. . a promise that that the sequence "continues in a similar manner",
. . that there is indeed a "reasonable" pattern to the terms.

I've posted the following before, but I enjoy showing it off.

Find the next term of the sequence: . $\bf{1,\,3,\,5,\,7,\,\cdots}$

Answer: 8

Justification
I was using this function: . $f(n) \;=\;-\frac{1}{24}\left(n^4 - 10n^3 + 35n^2 - 98x + 48\right)$

"That's unfair!" you say?
Okay, here's a better explanation . . .

The sequence: . $1,\,3,\,5,\,7,\,8,\,\cdots$ is the sequence of natural numbers
. . whose English names contain the letter "e".

My favourite is: find the next term in 1, 2, 4, 8, 16.

The answer that I am thinking of is 31. See here in particular scroll down to sequence A000127.

RonL