@Arjun Sir I think we can also develop the recurrence like :
T(n) = T(n-1) + T(n-3) +1 ; otherwise
T(n) = 1 ; for n < 1
Which means if the node at which we are standing is greater than equal to one then we will calculate the calls in (n-1) part + in (n-3) part + 1 (since the node on which we are standing is also a call) and in case if n<1 then that will yields only 1.
T(1)=3 , T(2)= 3 + 1 + 1 = 5, T(3) = 5 + 1 + 1 = 7 , T(4) = 7 + 3 + 1= 11 , T(5) = 11 + 5 + 1 = 17,
T(6) = T(5) + T(3) + 1 = 17 + 7 + 1 = 25.