Problem 16 from Section 2.3:

Let's say that a herd of female cows splits up into three types, female calves (in the year of birth), female yearlings (in year 1), and the mature females (all the rest of the females). According to the way the book states the problem, a female cow is not regarded as an adult until it gives birth, but I would call such a cow a mother cow. Since every mother has exactly one female calf each year, the number of mother cows in the herd each year is the same as the number of female calves (except for the first year, since we start with a calf whose mother was not in the herd). Thus we don't need to do anything special to keep track of the number of adult (i.e., mother) cows since we're already going to keep track of the number of calves.

The reproduction rule that we'll use for female cows is:
• They are immortal
• Each mature female gives birth the next year to a single female calf
• After a year, a calf becomes a yearling, and one year after that the yearling becomes a mature cow

Let's make tables showing what happens to the three classes of female cows in successive years:

The first three tables show the original calf as it grows up (note how the number 1 moves diagonally down the table):

Year 1:Years 1 and 2:Years 1, 2, and 3:
Year n 1 1 0 0 1
Year n Cn=# female calves in year n 1 2 1 0 0 1 0 0 1 1
Year n Cn=# female calves in year n Yn=# female yearlings in year n 1 2 3 1 0 0 0 1 0 0 0 1 1 1 1

The next three tables show the calf, now grown up, giving birth each year and its daughters growing up:

Up to Year 4:Up to Year 5:Up to Year 6:
Year n Cn=# female calves in year n Yn=# female yearlings in year n Mn=# mature females in year n 1 2 3 4 1 0 0 1 0 1 0 0 0 0 1 1 1 1 1 2
Year n Cn=# female calves in year n Yn=# female yearlings in year n Mn=# mature females in year n 1 2 3 4 5 1 0 0 1 1 0 1 0 0 1 0 0 1 1 1 1 1 1 2 3
Year n Cn=# female calves in year n Yn=# female yearlings in year n Mn=# mature females in year n Tn=total # female cows in year n 1 2 3 4 5 6 1 0 0 1 1 1 0 1 0 0 1 1 0 0 1 1 1 2 1 1 1 2 3 4

Things continue in the same way, except the grown up daughters now start having calves. In particular, the next table shows how to fill in the data for year n+1 given the data for year n.

 n n+1 Cn Cn+1=Mn Yn Yn+1=Cn Mn Mn+1=Mn+Yn Tn Tn+1

Here is the table through year 11:

 Year n Cn=# female calves in year n Yn=# female yearlings in year n Mn=# mature females in year n Tn=total # female cows in year n 1 2 3 4 5 6 7 8 9 10 11 1 0 0 1 1 1 2 3 4 6 9 0 1 0 0 1 1 1 2 3 4 6 0 0 1 1 1 2 3 4 6 9 13 1 1 1 2 3 4 6 9 13 19 28

Looking at the table we can see a pattern: the total next year is the sum of the totals this year and two years ago; i.e.,
Tn+1 = Tn + Tn-2.
Let's see if we can figure out why this must be true.
We know the total in any given year is just the sum of the calves, yearlings and mature cows; i.e.,
Tn+1 = Cn+1 + Yn+1 + Mn+1.
But the calves in a given year come from the mature cows of the previous year; i.e.,
Cn+1 = Mn.
Also, the yearlings in a given year come from the calves the previous year; i.e.,
Yn+1 = Cn,
and the mature cows in a given year come from the mature cows and the yearlings the previous year; i.e.,
Mn+1 = Mn + Yn.
Thus changing Cn+1 to Mn, and Yn+1 to Cn and Mn+1 to Mn + Yn gives
Tn+1 = Cn+1 + Yn+1 + Mn+1 = Mn + Cn + (Mn + Yn)
or
Tn+1 = Mn + Cn + (Mn + Yn) = (Mn + Cn + Yn) + Mn = Tn + Mn.
But in the same way,
Mn = Mn-1 + Yn-1 = (Mn-2 + Yn-2) + Cn-2 = Tn-2
so substituting Mn = Tn-2 into Tn+1 = Tn + Mn gives Tn+1 = Tn + Tn-2. Now we can answer parts (a) through (d) of problem 16:
(a) The book in this part of the problem is asking for the total number of cows and calves, but from context I believe that by this the book means just our Tn; i.e., the total number of cows of all types for years 1 through 11, which is just the bottom row in the table above.
(b) As explained above, the number of "adult" cows (using the book's terminology) for years 1 through 11 is just the number of mother cows, and since each mother has exactly one calf each year, this is given by Cn, except in year n = 1, since the calf we started with did not have a mother in our herd. For year n = 1, there were no mother cows in our herd.
(c) The number of calves each year is given by Cn.
(d) The recursion formula is Tn+1 = Tn + Tn-2. Thus T12 = T11 + T9 = 28 + 13 = 41, and likewise T13 = 60, T14 = 88, T15 = 129 and T16 = 189. (Note that there are other recursion formulas that work here too, so my answer is not the only possible answer.
For example, Tn+1 = Tn-1 + Tn-2 + Tn-3 is true. One way to justify this is to start with the recursion formula
Tn+1 = Tn + Tn-2. This also means Tn = Tn-1 + Tn-3, so we get
Tn+1 = (Tn-1 + Tn-3) + Tn-2.