# Series: Find the missing # in the series: 65, 33, _, 9

#### shastri.chandoo

##### New member
Find the missing # in the series: 65, 33, _, 9

#### MarkFL

##### Super Moderator
Staff member
I would guess the $$n$$th term in the series is:

$$\displaystyle a_n=2^{7-n}+1$$

Can you proceed?

#### tkhunny

##### Moderator
Staff member
Too few data points to pin it down. You need infinitely many. Three is no good and ANY answer would only be guessing.

I'm pretty fond of 43/3. -- Nice, level second differences.

Of course, many such sequences are not mathematical at all. Maybe it's the number of overdrawn accounts at the 14th street credit union on the west side of Wyomissing, PA.

Unless you can provide additional information, that's sort of where we are.

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#### Dr.Peterson

##### Elite Member
I would make the same GUESS as MarkFL, because it makes some nice patterns: each number is 1 more than a (descending) power of 2; the differences are descending powers of 2.

But any number could be put there, and it would still be a sequence. (Series is not really the right word.)

A question like this is not really a math problem, but a psychological test: do you see the same shape in this ink blot as the author did? People familiar with certain types of math are likely to see the same thing, but that doesn't really make it more correct than any other.

Thank you.

#### Jomo

##### Elite Member
As pointed out already, 17 is one of many correct answers. There are for example MANY polynomials that pass through the points (1,65),(2, 33) and (4, 9). Any one of those formulas (polynomial equations) would work to find the missing number.

#### Denis

##### Senior Member
Gets kinda weird using Mark's formula:
65, 33, 17 ,9 ,5 ,3 ,2 , 3/2, 5/4, 9/8, 17/16, ....

#### Jomo

##### Elite Member
Gets kinda weird using Mark's formula:
65, 33, 17 ,9 ,5 ,3 ,2 , 3/2, 5/4, 9/8, 17/16, ....
Mark, don't worry as things are always weird looking for Denise.
I myself prefer to think that an+1= (an+1)/2

#### MarkFL

##### Super Moderator
Staff member
I would probably choose to express the recursion in homogeneous form:

$$\displaystyle 2a_{n+2}-3a_{n+1}+a_{n}=0$$

Then it's easy to see that the characteristic roots are:

$$\displaystyle r\in\left\{\frac{1}{2},1\right\}$$

And so the closed-form will be:

$$\displaystyle a_n=c_12^{-n}+c_2$$

Then we use known values to determine the parameters:

$$\displaystyle a_1=c_12^{-1}+c_2=65\implies c_1+2c_2=130$$

$$\displaystyle a_2=c_12^{-2}+c_2=33\implies c_1+4c_2=132$$

Solving this system, we find:

$$\displaystyle \left(c_1,c_2\right)=\left(2^7,1\right)$$

Hence:

$$\displaystyle a_n=2^7\cdot2^{-n}+1=2^{7-n}+1$$

Thank you.

#### Denis

##### Senior Member
That is a plausible answer out of an infinitude of possible answers.
infinitude
[in-fin-i-tood, -tyood]
noun infinity: divine
an infinite extent, amount, or number.

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#### Jomo

##### Elite Member
Try an = (20/3)n^2 - 52n + 331/3

#### Denis

##### Senior Member
Try an = (20/3)n^2 - 52n + 331/3
That gives this pattern:
n
1-3: 65, 33, 14 1/3
4-6: 9, 17, 38 1/3
7-9: 73, 121, 182 1/3
10-12: 257, 344, 446 1/3
13-15: 561, 689, 830 1/3
and so on...

#### Jomo

##### Elite Member
infinitude
[in-fin-i-tood, -tyood]
noun infinity: divine
an infinite extent, amount, or number.
You finally learned how to use a dictionary. Good for you! You have come so far since coming to this forum. I think that you are ready to solo, SO GO AWAY!