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### Counting Bug Populations

```
Date: 12/03/98 at 09:04:16
From: Naveed Ahmed
Subject: Probability, sequences, formulas

We are given the following information:

A happy bug splits into a sad bug and a blank bug.
A sad bug splits into 2 happy bugs.
A blank bug splits into a sad bug and a happy bug.

The generation starts off with 1 happy bug which splits and forms
the 2nd generation.

The bugs die when they split.
The bugs live the same amount of time.

I have worked the first five generations to find how many of each kind
of bugs there are, but I can't find the formula for the nth generation.
Could also you tell me the formulas if the first generation starts with
a sad bug and a blank bug?

Thanks.
```

```
Date: 12/03/98 at 11:39:23
From: Doctor Anthony
Subject: Re: Probability, sequences, formulas

The first step is to make a table, using the following notation:
n = generation, H = happy bug, S = sad bug, and B = blank bug.

n   1   2   3   4    5    6    7    8     9   10     11    12
----------------------------------------------------------------
H   1   0   2   2    6   10   22   42    86   170   342   682
B   0   1   1   3    5   11   21   43    85   171   341   683
S   0   1   1   3    5   11   21   43    85   171   341   683
-----------------------------------------------------------------
Total 1   2  2^2 2^3  2^4 2^5  2^6   2^7  2^8   2^9   2^10  2^11

The pattern shows H is one more and one less alternatively than B and
S as we move up the generations. The total for the nth generation is
always given by 2^(n-1). Also each of H, B and S is close to 1/3 of
the total for any generation. To take account of the pattern we can
give two formulae depending on whether n is even or odd. A little
juggling shows the following is the pattern:

n EVEN                      n ODD
---------                 -----------

2^(n-1)- 2                2^(n-1)+ 2
H =  -----------          H  = ------------
3                          3

2^(n-1)+ 1                2^(n-1)- 1
B and S = ----------      B and S = ------------
3                         3

If you start with a first generation of S + B, you get the same table
as above except that n-1 is replaced by n. So the new formulae can be
produced by using n instead of n-1 and swapping over the n = even and
n = odd labels:

n ODD                   n EVEN
-----------            ---------------

2^n - 2                  2^n + 2
H = ---------           H =  ----------
3                        3

2^n + 1                 2^n - 1
B and S =  --------     B and S = ---------
3                      3

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Discrete Mathematics
High School Sequences, Series

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