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What number comes next and why?

 
 
Reply Mon 19 Jun, 2006 04:28 pm
1-3-5-9-15-31-61-125-251-503-1015-2035-4081-8177-16367-32747-???
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Type: Discussion • Score: 1 • Views: 6,348 • Replies: 13
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Tryagain
 
  1  
Reply Mon 19 Jun, 2006 05:49 pm
The answer to this is 65511


Why indeed, I will wait to see if anyone (Mark) agrees Laughing
0 Replies
 
markr
 
  1  
Reply Tue 20 Jun, 2006 12:24 am
I don't have a clue. Embarrassed
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Tryagain
 
  1  
Reply Tue 4 Jul, 2006 01:35 pm
Whoops, I almost forgot about this.


It's complicated to detail the method for achieving the answer, so bear with me please...

Let n be the position in the sequence, and R(n) be the answer - so in this case R(1)=1, R(2)=3, R(3)=5 etc.

Also, let X(n) be the sequence of prime numbers: X(n)=1, 2, 3, 5, 7, 11, 13, 17 etc

Okay - let Y(n) be X(n+1) - X(n).
So Y(n) = 1, 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6 etc.

And thus Y(n)-2 = -1, -1, 0, 0, 2, 0 , 2, 0 , 2, 4 etc

Now let D(n) be the SUM of ALL {Y(n)-2} from n=0 to n

i.e. the sequence D(n) = -1, -2, -2, -2, 0, 0, 2, 2, 4, 8 etc.

The solution is thus:
R(n) = 2^(n-1) - D(n) - 1.

So, when n=6, 2^(6-1) = 32; D(n) = 0; so R(6) = 32 - 0 - 1 = 31.

When n=12, 2^(12-11) = 2048; D(12) = 12; so R(12) = 2048 - 12 - 1 = 2035.

Likewise, when n=17 (i.e. the next in the sequence):
2(n-1) = 2^(17-1) = 65536; D(17) = 24; so R(17) = 65536 - 24 - 1 = 65511.


The only difficulty is in working out D(n). However, I'll leave that to you.
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markr
 
  1  
Reply Tue 4 Jul, 2006 10:12 pm
"Also, let X(n) be the sequence of prime numbers: X(n)=1, 2, 3, 5, 7, 11, 13, 17 etc"

But one isn't prime. Where in hell did you come up with that?
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Vi
 
  1  
Reply Wed 5 Jul, 2006 10:53 am
I am completely baffled as to how anyone could ever figure that out.

Here is a link to the online dictionary of integer sequences: http://www.research.att.com/~njas/sequences/

it's always surprising to see a sequence not included in the dictionary, so you should send it in.
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markr
 
  1  
Reply Wed 5 Jul, 2006 12:05 pm
This one seems rather contrived.
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whimsical
 
  1  
Reply Wed 12 Jul, 2006 02:59 am
How about this one. What should come logically on the question marks


[1/2 1/2] [1/4 1/2 1/4] [ ??? ] [1/8 1/4 1/4 1/4 1/8]
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markr
 
  1  
Reply Wed 12 Jul, 2006 10:31 am
How about
1/6 1/3 1/3 1/6
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lennylamus
 
  1  
Reply Fri 21 Jul, 2006 08:15 pm
The Fibbonacci
Haven't you guess seen the Da Vinci Code? It talks about teh Fibbonacci Sequence. All you have to do is do it in a different matter though. So the answer is...I don't know, because that isn't the fibbonacci. Oh well. And i hate fractions. I don't like them, well, except for some things. Just not this.
0 Replies
 
Vi
 
  1  
Reply Fri 21 Jul, 2006 10:21 pm
or maybe [1/8 1/8 1/2 1/8 1/8]
0 Replies
 
Junior86
 
  1  
Reply Wed 9 Aug, 2006 11:03 am
Tryagain wrote:
Whoops, I almost forgot about this.


It's complicated to detail the method for achieving the answer, so bear with me please...

Let n be the position in the sequence, and R(n) be the answer - so in this case R(1)=1, R(2)=3, R(3)=5 etc.

Also, let X(n) be the sequence of prime numbers: X(n)=1, 2, 3, 5, 7, 11, 13, 17 etc

Okay - let Y(n) be X(n+1) - X(n).
So Y(n) = 1, 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6 etc.

And thus Y(n)-2 = -1, -1, 0, 0, 2, 0 , 2, 0 , 2, 4 etc

Now let D(n) be the SUM of ALL {Y(n)-2} from n=0 to n

i.e. the sequence D(n) = -1, -2, -2, -2, 0, 0, 2, 2, 4, 8 etc.

The solution is thus:
R(n) = 2^(n-1) - D(n) - 1.

So, when n=6, 2^(6-1) = 32; D(n) = 0; so R(6) = 32 - 0 - 1 = 31.

When n=12, 2^(12-11) = 2048; D(12) = 12; so R(12) = 2048 - 12 - 1 = 2035.

Likewise, when n=17 (i.e. the next in the sequence):
2(n-1) = 2^(17-1) = 65536; D(17) = 24; so R(17) = 65536 - 24 - 1 = 65511.


The only difficulty is in working out D(n). However, I'll leave that to you.


genius!...

so is it correct then?
0 Replies
 
lilarban98
 
  1  
Reply Thu 10 Aug, 2006 03:16 pm
ooooo oooo i like pokemon i like pokemon!!!!!!!!!
0 Replies
 
misogi
 
  1  
Reply Wed 23 Aug, 2006 06:06 am
algorithms... yawn.
0 Replies
 
 

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