Thread: sequence
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Old 2004-04-04, 05:47   #4
Kevin
 
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Aug 2002
Ann Arbor, MI

433 Posts
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I can get it down to a recursive of sorts....

x[n]=(1+x[0]^3+x[1]^3...+x[n-1]^3)/n
n*x[n]=(1+x[0]^3+x[1]^3...+x[n-2]^3)+x[n-1]^3
n*x[n]-x[n-1]^3=(1+x[0]^3+x[1]^3...+x[n-2]^3)
(n*x[n]-x[n-1]^3)/(n-1)=(1+x[0]^3+x[1]^3...+x[n-2]^3)/(n-1)
(n*x[n]-x[n-1]^3)/(n-1)=x[n-1]
n*x[n]-x[n-1]^3=(n-1)*x[n-1]
x[n]={(n-1)*x[n-1]+x[n-1]^3}/n

edit: lil bit more
x[n]=x[n-1]*[n-1+x[n-1]^2]/n
x[n]=x[n-1]*{n-(x[n-1]-1)(x[n-1]+1)}/n
this tells us that if n divides x[n-1]-1, x[n-1], or x[n-1]+1, then x[n] is integral. If it doesn't, then x[n] is not an integer.

Last fiddled with by Kevin on 2004-04-04 at 05:54
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