Finding recurrence relation for variation on Fibonacci sequence

Question

(a) A young pair of rabbits is placed on an island. A young pair does not breed until they are 2 months old. After they are 2 months old, each pair of rabbits produces another pair each month. Assuming that no rabbit ever dies, find the number of rabbits living in the island after n months. Find the recurrence relation.

(b) Try to solve the following variants of the problem: (i) when a young pair does not breed until they are 3 months old, (ii) when a young pair does not breed until they are N months old.

Solution of (a) is Fibonacci equation ${f_n} = {f_{n-1}}+{f_{n-2}}$ for $n\ge2$.

For (b)(i) i got a sequence of 1,1,1,2,3,4,6,9, ...

But I got stuck in finding a recurrence relation for it and for (ii).

Answer 1

Yes, in the (b) i) case you get the equation $$ f_0=f=1=f_2=1\ ;\ f_n=f_{n-1}+f_{n-3}\ (n\geq 3) $$ then, all depends what is required. Is it (1) a closed form ? (2) the generating function $F=\sum_{n\geq 0}f_n\,z^n$ ?

Likewise, for the (b) ii) case, you get $$ f_0=f=1=f_2=\cdots =f_{N-1}=1\ ;\ f_n=f_{n-1}+f_{n-N}\ (n\geq N) $$

Answer 2

I believe $f_x=f_{x-1}+f_{x-n-1}$ works (for $x>n+1$) and $f_1=f_2=\cdots=f_{n}=1 $, for the generalised (ii).

Reasoning:

There are $f_{x-1}$ rabbits that were already alive and $f_{x-n-1}$ rabbits are able to breed by the xth month.