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| | #1 |
| Registered User Join Date: Aug 2003
Posts: 782
| Number sequence help I know that a(1) = 2 and I know that a(2) = 3. Now I need to find an explicit formula for that. I know it has something to do with fibonaccis sequence since, if a(1)=1 and a(2) = 1 the explicit formula would have been a(n) = (2^n-(-1)^n)/3. The problem that this does not work when a(1) = 2 and a(2) = 3. Can anybody explain how I should come up with an explicit formula for this problem? |
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| | #2 |
| 5|-|1+|-|34|) Join Date: Aug 2001
Posts: 4,429
| The answer..... is 42. |
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| | #3 |
| Registered User Join Date: Aug 2003
Posts: 782
| herrrrrrrrrrrm.......ummmmmmmmm.......no |
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| | #4 |
| and the hat of marbles Join Date: May 2002 Location: Lund, Sweden
Posts: 2,041
| It can be solved with matrices. I'll see if I can produce some images for you.
__________________ Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling |
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| | #5 |
| Registered User Join Date: Aug 2003
Posts: 782
| That would be very nice!! |
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| | #6 |
| and the hat of marbles Join Date: May 2002 Location: Lund, Sweden
Posts: 2,041
| OK, the board is semi-down right now, so we'll see if this gets through.. I'm going to assume that you know how to handle matrices. First, we'll define a matrix A like this:
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| | #7 |
| and the hat of marbles Join Date: May 2002 Location: Lund, Sweden
Posts: 2,041
| Ahh, the boards are back! Now, we add your recursive formula:
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| | #8 |
| Registered User Join Date: Aug 2003
Posts: 782
| Ok....yah I know how to handle matrices (at least to the point where I can multiply them). |
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| | #9 |
| and the hat of marbles Join Date: May 2002 Location: Lund, Sweden
Posts: 2,041
| We want to diagonalize X, to make it easy to calculate X^n.
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| | #10 |
| and the hat of marbles Join Date: May 2002 Location: Lund, Sweden
Posts: 2,041
| When we perform the last multiplication with A2, the first row of the result will be a(n+2). Here's the final formula:
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| | #11 |
| and the hat of marbles Join Date: May 2002 Location: Lund, Sweden
Posts: 2,041
| Numbers generated from this formula: 3 7 13 27 53 107 213 427 853 1707 3413 6827 13653 27307 54613 109227 218453 436907 873813 1747627 3495253 Seems correct for the first three, at least. HTH
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| | #12 |
| Registered User Join Date: Aug 2003
Posts: 782
| That definately seems to be correct! Thanks a bunch, now I just have to get a good grip on everything but that shouldnt be too hard... ! Thanks.It is correct for the 18 first numbers. Last edited by Shakti; 05-26-2005 at 08:52 AM. |
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