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New Member
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Dec 2, 2009, 12:09 PM
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1,2,6,? 72
What are the two numbers missing in the following sequence of numbers : 1,2,6,? 72
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Uber Member
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Dec 2, 2009, 06:26 PM
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Hi, jk!
Are these homework questions that you've been asking on this site?
Thanks!
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Junior Member
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Dec 2, 2009, 09:58 PM
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Hi,jk and Clough.
1,2,6,? 72
The answer might be 1,2,6,12,36,72.
But I think the question was not asked in a right way. jk,You should give your question more clearly. On the other hand, I think some homeworks should be solved by yourself, and if you are confused at it then ask here.
Thanks!
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Uber Member
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Dec 3, 2009, 10:57 AM
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I can't find another possible sequence... :(
If that was like that: (1, 2), (6, ), (? 72) then I'd have put what shihouzhuge has put.
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Junior Member
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Dec 4, 2009, 02:55 AM
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Originally Posted by Unknown008
I can't find another possible sequence... :(
If that was like that: (1, 2), (6, ?), (?, 72) then I'd have put what shihouzhuge has put.
Unknown008.
:), I think the regular may be *2,*3,*2,*3...
Thanks!
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Junior Member
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Dec 4, 2009, 06:37 AM
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Tn = T(n-1) + 3T(n-2) +1
gives
1, 2, 6, 13, 32, 72
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Junior Member
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Dec 4, 2009, 07:12 AM
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Originally Posted by elscarta
Tn = T(n-1) + 3T(n-2) +1
gives
1, 2, 6, 13, 32, 72
Hi,elscarta.
I think you give a nice answer, and the question may have many answers.
Good job!
Thanks!
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Junior Member
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Dec 4, 2009, 07:48 AM
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Mathematically it is possible to find a polynomial of order 5 which gives any numbers that you want for the missing numbers.
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Uber Member
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Dec 4, 2009, 10:02 AM
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Originally Posted by elscarta
Tn = T(n-1) + 3T(n-2) +1
gives
1, 2, 6, 13, 32, 72
I would never have thought of that one... You used some program?
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Junior Member
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Dec 4, 2009, 05:14 PM
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Originally Posted by Unknown008
I would never have thought of that one... You used some program?
Actually it was trial and error.
Firstly I noticed that 2 x (2 + 1) = 6 so maybe T(n) = 2x(T(n-1) + T(n-2))
but this gives
1, 2, 6, 16, 44, 120 which is too big so I reduced the weight of the T(n-1) term
So I thought 2 + 4 x 1 = 6 so maybe T(n) = T(n-1) + 4x T(n-2) but this gives
1, 2, 6, 14, 38, 94 which is still too big so I thought that there must be a constant added in not just a combination of the previous terms.
This lead to 2 + 3 x 1 + 1 = 6 so maybe T(n) = T(n-1) + 3x T(n-2) +1 which turned out to work.
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Uber Member
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Dec 4, 2009, 10:49 PM
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Wonderful! Genius! :)
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