# Math Help - Number pattern.

1. ## Number pattern.

1,1,2,4,16,128,4096

What are the next 3 terms.

2. the next $2$ are $1048576$ and $8589934592$ I will let someone else reveal the third

3. Next 3:

18014398509481984, 309485009821345068724781056, 11150372599265311570767859136324180752990208

4. I would have thought that would have taken some time. It would have for me .

Please tell me how you can find patterns so quick? Or is it just, your that good at pattern recognition? Thanks

5. Originally Posted by integral

Please tell me how you can find patterns so quick?
This one was quite easy. This is what to do. Write the numbers down and start looking at any relationships between consecutive numbers.

In this case look at the ratios between each consecutive numbers. For the next term consider the relationship between the previous two ratios.

$1 {\color{red}\times 1 = } , 1 {\color{red}\times 2 = } , 2 {\color{red}\times 2 = } , 4 {\color{red}\times 4 = } , 16 {\color{red}\times 8 = } , 128 {\color{red} \times 32 = } ,\dots$

6. Originally Posted by integral
I would have thought that would have taken some time. It would have for me .

Please tell me how you can find patterns so quick? Or is it just, your that good at pattern recognition? Thanks
The powers of two jumped out at me pretty quickly.

But just look at each term divided by the preceding term and you should see the pattern quickly too.

Edit: Forgot to refresh the page before posting, so my "hint" is pretty unnecessary after pickslides's post.

7. Originally Posted by pickslides
This one was quite easy. This is what to do. Write the numbers down and start looking at any relationships between consecutive numbers.

In this case look at the ratios between each consecutive numbers. For the next term consider the relationship between the previous two ratios.

$1 {\color{red}\times 1 = } , 1 {\color{red}\times 2 = } , 2 {\color{red}\times 2 = } , 4 {\color{red}\times 4 = } , 16 {\color{red}\times 8 = } , 128 {\color{red} \times 32 = } ,\dots$
Ohrly? Interesting.

I was playing around to find a number pattern so I started with pascals triangle.
$
\binom{n}{k}$

then I added up each row, but not all rows. I chose a row correlating to a financing number subtracted by 1.

so

$
\binom{(F_{n}+F_{n-1})-1}{k}$

which gives

$\binom{1-1}{k}$
$\binom{1-1}{k}$
$\binom{2-1}{k}$
$\binom{3-1}{k}$

ect.

But I wanted the sum so.

$\sum^{F_n-1}_{k}\binom{F_n-1}{k}$
$\left [ \sum^{0}_{k}\binom{0}{k} \right ]=1$
$\left [ \sum^{0}_{k}\binom{0}{k} \right ]=1$
$\left [\sum^{1}_{k}\binom{1}{k}\right ]=2$
$\left [ \sum^{2}_{k}\binom{2}{k}\right ]=4$
$\left [ \sum^{4}_{k}\binom{4}{k}\right ]=16$
$\left [ \sum^{7}_{k}\binom{7}{k}\right ]=128$

While your method works yes (although I did not think of it)

It does not explain the first term, only all after.

1,1,2,4,16

So, you're answers work but are technically incorrect.

$
\int$

8. Originally Posted by integral
So, you're answers work but are technically incorrect.
I think the answers are just as correct as yours.

The pattern stated in pickslides's post requires the first three values to be given, just like how the Fibonacci sequence requires the first two values, as base cases for recursion.

9. Originally Posted by integral
It does not explain the first term, only all after.
No need to; the 1st term here is really the "2"; the two 1's are to set up;
same as with fibonacci sequence's set up with 0 and 1.

10. I suppose, indeed.

Interesting non the less

11. Look at the power of the pattern

$0,0,1,2,4,7,12 ....$

If we add one in each term

we have

$1 , 1 , 2,3,5,8,13 ...$

Fibonacci sequence

12. Originally Posted by simplependulum
Look at the power of the pattern

$0,0,1,2,4,7,12 ....$

If we add one in each term

we have

$1 , 1 , 2,3,5,8,13 ...$

Fibonacci sequence
What is the pattern of? (the first one not the fibonacci.)

13. Originally Posted by integral
What is the pattern of? (the first one not the fibonacci.)
Powers of 2.

$2^0 = 1$
$2^0 = 1$
$2^1 = 2$
$2^2 = 4$
$2^4 = 16$
$2^7 = 128$

etc.

14. I see, so any sequence you place at the top of your binomial coefficient will be two to the power of that sequence?

awesome!