Harriss: Rabbit Sequence

There has been a theme in some of the recent posts and problems. It’s a little buried but almost enough to say its another of those Mathfactor agendas when we try to sneak some knowledge to you buried in the fun.  Never one to miss such an opportunity I will jump in with a post, and a problem.  This is a slight change to a classic problem that comes out of the work of one of my mathematical heroes:  Leonardo of Pisa, also known as Fibonacci.  He is responsible for changing how we count! Not many people can claim that. He introduced the system base value, also known as Arabic numerals that we still use today into Europe.  He is more famous however for talking about rabbits:

Imagine that you have immortal rabbits, Bugs Bunny’s version of Olympus perhaps.  Even if they are immortal however rabbits are famous for one thing.  They breed like, well rabbits.  Some of the rabbits are children and some adults and are divided into pairs.  Each month any child pairs become adults and any adult pairs breed to produce a new child pair.  They are immortal so no pair ever dies.  These rabbits are also a little odd.  They live on a line (don’t complain, this is no more ludicrous than that they are immortal!), but can shuffle along.  Also if you are worried about inbreeding, the male rabbit leave the family hutch and shuffle along the line past others until they find a suitably unrelated mate.  Why we would be worried about inbreeding in immortal rabbits living on a line escapes me!

Anyway we start with one pair of children.  Lets put a c.  After a month they become adults, a.  Another month passes and they now have a pair of children, but are still there themselves.  We therfore have the original pair and a pair of children: ac.  Next month the adults have another pair of children and the children become adults: aca.  Can you see how this will work?  Each month the children become adults so we replace every c with and a, each pair of adults has a new pair of children but stays as adults, so we replace every a with ac.  We can continue to get longer and longer sequences of rabbits on this line:

aca  to acaac to acaacaca to acaacacaacaac….

Now some puzzles.  Given a line with 21 adult pairs and 13 child pairs, how many pairs of adults and children would there be after one month?

Given p adults and q children how many adults and children will there be after one month?

Finally a more difficult one.  How will the ratio of adults to children behave month on month?  Will it

a) Get closer and closer to a particular number?

b) Keep on changing without pattern?

In either case can you say more?


  1. mathphan said,

    April 1, 2009 at 10:42 am

    The clues are all there…[spoiler]This is directly related to Fibonacci sequences.  If you just write the number of adults and children at each stage you have:
    a c
    0 1
    1 0
    1 1
    2 1
    3 2
    5 3
    8 5
    13 8
    21 13
    34 21

    The number of adults is equal to the number of adults in the prior month plus the number of children in the prior month (that mature to be adults).
    a(n) = a(n-1) + c(n – 1)

    Also, the number of children is just the number of adults in the prior month (because they all have offspring).
    c(n) = a(n – 1)

    Putting the two equations together you have:
    a(n) = a(n – 1) + a(n – 2)

    This is the classic Fibonacci sequence.

    To answer the questions:
    1) (21, 13) is followed by (34, 21)
    2) (p, q) is followed by (p+q, p)
    3) See below:

    Let’s compare the ratios of adults to children in successive months.
    p/q compared to (p+q)/p

    Let’s say this does converge on a particular ratio.  I’ll call that ratio R.
    R = p/q = (p+q)/p

    Split the fraction at the end:
    R = p/q = p/p + q/p
    R = p/q = 1 + q/p

    Substituting in 1/R for q/p:
    R = 1 + 1/R

    Multiply both sides by R:
    R² = R + 1

    Get everything on the left hand side:
    R² – R – 1 = 0

    This can be solved with the quadratic formula or completing the square.
    R² – R  = 1
    R² – R + 1/4 = 5/4
    (R – 1/2)² = 5/4
    (2R/2 – 1/2)² = 5/4
    (2R – 1)²/4 = 5/4
    (2R – 1)² = 5
    2R – 1 = ± ?5
    2R = 1 ± ?5
    R = (1 ± ?5) / 2

    In our example the ratio is always greater than 1 (more adults than children so we want:
    R = (1 + ?5) / 2

    That is the Golden Ratio:
    R ? 1.618033988749894848204586834…[/spoiler]

  2. mathphan said,

    April 1, 2009 at 10:47 am

    For anyone that is wondering, a couple characters in my prior response got replaced. [spoiler]?5 is sqrt(5).  The square root symbol got changed.
    Also, I used the symbol for approximately equal when giving the decimal expansion of phi.[/spoiler]

  3. Gelada said,

    April 3, 2009 at 11:26 am

    Great answer mathphan!  You have essentially given my next post on this topic!

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