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 \(a_2,a_3,and\) \(a_4\)?

Havingfun Jun 19, 2019

#1**+2 **

I think there are 502 that satisfy the condition.

Only the odd numbers provide a_{2} greater than a_{1}. There are 1004 of them.

All these a_{2}'s are even, so a_{3} = a_{2}/2 = (3a_{1} + 1)/2 which is greater than a_{1}.

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

Alan Jun 19, 2019