What are the two numbers missing in the following sequence of numbers : 1,2,6,? 72

Hi, jk!

Are these homework questions that you've been asking on this site?

Thanks!

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!

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.

Quote:

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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.

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Unknown008.
:), I think the regular may be *2,*3,*2,*3...
Thanks!

Tn = T(n-1) + 3T(n-2) +1

gives

1, 2, 6, 13, 32, 72

Quote:

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Originally Posted by elscarta

Tn = T(n-1) + 3T(n-2) +1

gives

1, 2, 6, 13, 32, 72

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Hi,elscarta.
I think you give a nice answer, and the question may have many answers.
Good job!

Thanks!

Mathematically it is possible to find a polynomial of order 5 which gives any numbers that you want for the missing numbers.

Quote:

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Originally Posted by elscarta

Tn = T(n-1) + 3T(n-2) +1

gives

1, 2, 6, 13, 32, 72

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I would never have thought of that one... You used some program?

Quote:

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Originally Posted by Unknown008

I would never have thought of that one... You used some program?

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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.

Wonderful! Genius! :)