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I am trying to understand a twist on the Fibonacci bunnies scenario, where the bunnies die x generations after their birth (where x is a positive integer).

An example is shown here.

I understand the table half-way down the link showing the case of x=2 (rabbits die 2 years later) (the table has 5 columns, where column 1 = "Year", column 2 = "Pairs of Newborn Rabbits", column 3 = "Pairs of Newly Mature Rabbits", column 4= "Pairs of Really Mature Rabbits", column 5 = "Total").

According to this table, the total number of rabbits each year follows the sequence 1, 1, 2, 2, 3, 4, 5, 7, 9, ...

The problem is, if asked on an exam to show the case of x=10, I would need 13 columns (too many) in order to derive the last column (and hence solve the sequence of total bunnies per year).

My question is, is there a more efficient table to use so that one could derive the sequence of total bunnies per year for large values of k?

For instance, a table like this one would be helpful (in answer 2). But I think the numbers of total bunnies may be incorrect (because when I try to reapply this table for x=2, I do not get the sequence from the first link of 1, 1, 2, 2, 3, 4, 5, 7, 9, which I know is correct.)

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I don't think you can really do much better than that table, except maybe to remove the "dead rabbits" column. –  Eric Stucky Oct 1 '13 at 23:01
    
A better approach: rather than keeping track of the age of all your bunnies, you can instead track how many bunnies were born each year - this is just the difference between how many bunnies are alive one year and how many are alive the next - and suitably 'kill off' that many bunnies after $x$ years. In other words, your sequence will be of the form $B_n=B_{n-1}+B_{n-2}-(B_{n-x}-B_{n-x-1})$. –  Steven Stadnicki Oct 1 '13 at 23:25
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1 Answer 1

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If you say the rabbits need one year to mature, then bear for two years, then die, your table looks like this: $$\begin {array} {r|r|r|r|r|r} year&newborn&one year&two years&three years&total\\1&1&0&0&0&1\\2&0&1&0&0&1\\3&1&0&1&0&2\\4&1&1&0&1&3\\5&2&1&1&0&4\\6&2&2&1&1&6\\7&4&2&2&1&9\end{array}$$

Letting $A(n)$ be the number of newborns in year $n$, $B(n)$ the yearlings, $C(n)$ the two year olds, $D(n)$ the three year olds, and $E(n)$ the total, we have $$A(n)=B(n-1)+C(n-1)+D(n-1)\\B(n)=A(n-1),C(n)=B(n-1),D(n)=C(n-1)\\A(n)=A(n-2)+A(n-3)+A(n-4)\\E(n)=A(n)+B(n)+C(n)+D(n)\\E(n)=A(n)+A(n-1)+A(n-2)+A(n-3)$$

The third line shows you don't need all the other columns, as you have a recurrence for $A$ that doesn't depend on $B,C,D$. For longer lived rabbits, you will have more terms, but it will work the same. We can find the rate of growth of $A$ by the usual approach of assuming a geometric growth, getting $r^4=1+r+r^2$ with the largest positive root about $1.4656$

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Thanks Ross. That is a pretty interesting observation. I made a similar table for x=10, and saw that it still worked for E(n)=A(n)+A(n-1)+...+A(n-11). –  Stella J Oct 2 '13 at 17:51
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