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Initial Term Identification (Posted on 2015-07-21) Difficulty: 3 of 5
A sequence {X0, X1, X2, ...., Xn} is given by:
Xn(1 - Xn-1) = Xn-1

Find X0, given that: X2015 = 2015

No Solution Yet Submitted by K Sengupta    
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Solution electronically computed | Comment 1 of 3
x(n) - x(n)*x(n-1) = x(n-1)
x(n-1)*(1 + x(n)) = x(n)
x(n-1) = x(n)/(1+x(n))

So

    4    kill "initterm.txt"
    5    open "initterm.txt" for output as #2
   10   X=2015:Lvl=2015
   20   for Lvl=2014 to 0 step -1
   30    X=X//(1+X)
   35   print #2,Lvl,X
   40   next
   45   close #2
   50   print X

In the first few terms the numerator remains 2015, but the denominator, initially 1, goes up by 2015 each successively smaller subscript

  n        x(n)
 2014   2015//2016 
 2013   2015//4031 
 2012   2015//6046 
 2011   2015//8061 
 2010   2015//10076 
 2009   2015//12091 
 2008   2015//14106 
 2007   2015//16121 
 2006   2015//18136 
 2005   2015//20151 
 
 
 ...
 
 and the last few are:
 
 ...
 
 9   2015//4042091 
 8   2015//4044106 
 7   2015//4046121 
 6   2015//4048136 
 5   2015//4050151 
 4   2015//4052166 
 3   2015//4054181 
 2   2015//4056196 
 1   2015//4058211 
 0   2015//4060226 
 
 So x(0) = 2015/4060226


  Posted by Charlie on 2015-07-21 15:08:06
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