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Number Sequences


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26 replies to this topic

#11 strumfika

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Posted 29 July 2007 - 01:55 PM

an interesting number sequence AHEM CUBA :
26,25,52,50,5,4,40,39... the next is 93 , but does 91 follow ?

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#12 snotty87

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Posted 27 August 2007 - 05:26 PM

For the 1,2,3,6,11,20,37,68 ... Someone didn't know the relationship between 1 and 2 ... recall this, it is a recursive equation which involves the three previous numbers, therefore the first three numbers are just given with no relationships between them!
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#13 BoilingOil

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Posted 29 September 2007 - 06:04 PM

* 8723, 3872, 2387, 7238
* 1, 4, 9, 18, 35, 68
* 23, 45, 89, 177, 353
* 7, 5, 8, 4, 9, 3, 10
* 11, 19, 14, 22, 17, 25, 20
* 3, 8, 15, 24, 35, 48
* 2, 4, 5, 10, 12, 24, 27, 54
* 1, 3, 4, 7, 11, 18, 29
* 99, 92, 86, 81, 77, 74
* 0, 4, 2, 6, 4, 8, 6
* 1, 2, 2, 4, 8, 11, 33, 37
* 1, 2, 6, 24, 120, 720
* 1, 2, 3, 6, 11, 20, 37, 68
* 5, 7, 12, 19, 31, 50, 81
* 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322
* 126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, 485
* 4, 7, 15, 29, 59, 117, 235
* 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2
* 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, 4


This is the kind of thing I trained my math skills with as a kid. Nice flashback

BoilingOil
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#14 BoilingOil

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Posted 29 September 2007 - 06:12 PM

• 1, 2, 2, 4, 8, 11, 33, ?

1 + 1 = 2
2 * 1 = 2
2 + 2 = 4
4 * 2 = 8
8 + 3 = 11
8 * 3 = 33
33 + 4 = 47
47 * 4 = 188
...



8 * 3 = 33 ????
33 + 4 = 47 ????

My memory might be failing me, because my basic math classes were over 40 years ago. Or maybe someone discovered new relationship between numbers. But I don't think so

Of course you meant: 11 * 3 = 33, 33 + 4 = 37
and the last line would then be 37 * 4 = 148

At least, if the old maths I learned are still valid

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#15 BoilingOil

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Posted 29 September 2007 - 06:18 PM

1,4,9,18,35,.....68
this was a reply I wrote down but cant remember how i got it, I jotted down 35+33=68 off to the side but I cant figure why.....hmmmm oh i got it 1 (2 spaces) 4 (4 spaces) 9 (8 spaces) 18 (16 spaces) 35 (32 spaces) 68 (64 spaces) 133 etc. etc.


23,45,89,177,.....303 23(+22) 45 (+44) 89(+88) 177(+176) 303(+352) 655 etc. etc.
I guess this proves there is more than one possible answer



177(+176) = 303 ????
Try that again, please. It should be 353
Ofcourse, the next one will then be wrong as well. It should read 353(+352) = 705


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#16 kingmadmushroom

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Posted 14 October 2007 - 09:34 AM

*

This is the kind of thing I trained my math skills with as a kid. Nice flashback

BoilingOil



yeah we r still doin this kinda thing now
i think they should put som harder ones on cos face it , im 12 an i can do all of these
so how easy must u guys find iit!!!!!!
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#17 coffee.lady

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Posted 06 November 2007 - 10:58 PM

1,2,3,6,11,20,37...68

a+b+c=d,b+c+d=e,c+d+e=f...
...I think?
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#18 Yog

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Posted 28 November 2007 - 05:49 PM

This is more satisfactory to me as a mathematical operation versus just rotating number order physically
Given 8723
subtract 4851
Given 3872
subtract 1485
Given 2387
Observing 4851 then 1485, the next subtraction number would be 5148
subtract 5148
ans: -3239

Comments?

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#19 otasyn

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Posted 28 November 2007 - 07:50 PM


• 1, 2, 2, 4, 8, 11, 33, ?

1 + 1 = 2
2 * 1 = 2
2 + 2 = 4
4 * 2 = 8
8 + 3 = 11
8 * 3 = 33
33 + 4 = 47
47 * 4 = 188
...



8 * 3 = 33 ????
33 + 4 = 47 ????

My memory might be failing me, because my basic math classes were over 40 years ago. Or maybe someone discovered new relationship between numbers. But I don't think so

Of course you meant: 11 * 3 = 33, 33 + 4 = 37
and the last line would then be 37 * 4 = 148

At least, if the old maths I learned are still valid

BoilingOil



According to sn00bino's attempt and BoilingOil correction, the 2nd unlisted # would be 148. However, I don't think that sequence is accurate. The numbers to the right of the = form the sequence, but what about the first number "1"? sn00bino almost has it. The correct sequence involves alternating "+" and "*", but the modifying values are not entirely correct. So, here is my opinion of the solution:

1 * 1 = 1
1 + 1 = 2
2 * 1 = 2
2 + 2 = 4
4 * 2 = 8
8 + 3 = 11
11 * 3 = 33
33 + 4 = 37
37 * 5 = 185

Here, we have alternating formulas (please think of the non-bold (n)'s and (n+1) as subscript counters instead of coefficients because I don't have actual subscript to work with):

a(n) * b(n) = c(n)
c(n) + n = a(n+1)

Every b(n) = Fibonacci(n). (ie 1,1,2,3,5,8...)

I should note that n does NOT represent a location in the sequence of single digits, but instead represents a location of paired digits:

For {1,2}, n = 1
For {2,4}, n = 2
For {8,11}, n = 3
For {33,37}, n = 4
For {185,190}, n = 5
And so on...

Of course, I think that sn00bino's and BoilingOil's observation were impressive nonetheless. Also, you could argue that their sequence is still correct by saying that the first digit "1" is just a starting point and the formula's do not start until the second digit.

Please correct me if I'm wrong, made a typo, and just plain don't agree.
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#20 spoxjox

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Posted 10 December 2007 - 11:55 PM

• 1, 2, 2, 4, 8, 11, 33, ?

1 + 1 = 2
2 * 1 = 2
2 + 2 = 4
4 * 2 = 8
8 + 3 = 11
8 * 3 = 33
33 + 4 = 47
47 * 4 = 188
...


Nicely done, though as someone already pointed out, 33 + 4 = 37, not 47, and 37 * 4 = 148, not 188. But nice job cracking it.

Another possibility: Include the inferred zero on the front, and you have, taking the numbers in successive (interlocking) pairs:

Start out with {0, 1}
+1 + (0 + 1) = 2 : (2 - 1) * 2 = 2 -- now we have {0, 1, 2, 2}
+0 + (2 + 2) = 4 : (4 - 2) * 4 = 8 -- now we have {0, 1, 2, 2, 4, 8}
-1 + (4 + = 11 : (11 - * 11 = 33 -- now we have {0, 1, 2, 2, 4, 8, 11, 33}
-2 + (11 + 33) = 42 : (42 - 33) * 42 = 378 -- now we have {0, 1, 2, 2, 4, 8, 11, 33, 42, 378}
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