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#1 |
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Dec 2002
2 Posts |
Hi Guys.
I hope this time to write better then the first time... ops: first question: What the next number of the sequence: 5 12 18 45 36 75 54 105 182 .... Ovviamente why is it the next. ;) second question: I need a dimonstration of this formula. I can't find no methods to explain this. ops: (x+1)^n= Sum from 1 to (n-1) of binomial coefficient from i to n-1 on (-1)^(n-1)*(x-i+1)^n+out of summatory (n!/2)*(2x-n+3) with n from 2 up. I thank all of us. Bye bye by Emanuele. |
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#2 | |
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"Mike"
Aug 2002
202A16 Posts |
Quote:
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#3 | |
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∂2ω=0
Sep 2002
República de California
19·613 Posts |
Quote:
a sequence, depending only on how complicated a formula one is willing to consider. But typically such problems ask if there is a simple formula (e.g. a 3-term recurrence) that generates the given sequence. There doesn't seem to be in this instance. Instead, let's look at the factorizations of these numbers: 5, 2^2.3, 2.3^2, 5.3^2, 2^2.3^2, 3.5^2, 2.3^3, 3.5.7, 2.7.13 Except for 5, these numbers are all smooth, in the sense that their largest prime factor is less than their square root. But if we were simply writing down integers that satisfy this property, in the order in which they occur, we'd get the following sequence: 8, 12, 16, 18, 24, 27, 32, 36, 40, 45, 48, 50, 54, 56, 60, 63, 64, 70, ... i.e. we get all the terms of the sequence (except 5), but lots more, too. OTOH, if you wanted to brute-force it, you could use the fact that there is an infinite number of polynomials of degree >= 9 which achieve the above set of y-values at any 9 distinct x-values, including the obvious progressions like x = 0,1,2,3,4,5,6,7,8. |
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#4 |
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Dec 2002
2 Posts |
I'm not so powerfull in math to think an impossible or a very difficult it's
more simple because every number have a similar type of production like the others and they are in sequence. the next is 90 what's the next? bye ;) |
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