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.