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Let \(a_1,a_2,\ldots\) be a sequence determined by the rule \(a_n= \frac{a_{n-1}}{2}\) if \(a_{n-1}\) is even and \(a_n=3a_{n-1}+1\) if \(a_{n-1}\) is odd. For how many positive integers \(a_1 \le 2008\) is it true that \(a_1\) is less than each of \(a_2,a_3,and\) \(a_4\)?

 Jun 19, 2019
 #1
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I think there are 502 that satisfy the condition.

 

Only the odd numbers provide a2 greater than a1.  There are 1004 of them. 

All these a2's are even, so a3 = a2/2 = (3a1 + 1)/2 which is greater than a1.

 

The corresponding a4's must alternate even and odd.  The even ones will be a3/2 = (3a1 + 1)/4 which must be less than or equal to a1 (equality when a1 = 1).  The odd ones must be   3(3a1 + 1)/2 + 1 = 9a1/2 + 5/2 > a1.  There are 1004/2 = 502 of these. 

 Jun 19, 2019

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