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Old 2003-02-06, 19:04   #1
Ema.abikyokan
 
Dec 2002

2 Posts
Default Two questions

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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Old 2003-02-06, 19:16   #2
Xyzzy
 
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"Mike"
Aug 2002

3·2,741 Posts
Default Re: Two questions

Quote:
Originally Posted by Ema.abikyokan
What the next number of the sequence:
5 12 18 45 36 75 54 105 182 ....
According to the On-Line Encyclopedia of Integer Sequences there is no pattern...
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Old 2003-02-07, 19:08   #3
ewmayer
2ω=0
 
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Sep 2002
República de California

103×113 Posts
Default Re: Two questions

Quote:
Originally Posted by Ema.abikyokan
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. ;)
Of course there could be any number of generating formulae for such
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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Old 2003-02-10, 17:26   #4
Ema.abikyokan
 
Dec 2002

2 Posts
Default little help for solution.

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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