A cow has a calf every year. A calf becomes a cow in 4 years. Starting with one cow, how many animals are there in 17 years?

Question

A cow gives birth to a calf every year. The calf becomes a cow in 4 years. The cow gives birth to a calf every year. Starting with one cow, how many animals are there in 17 years?

P.S. The cows live forever.

P.S. May have something to do with Narayana.

Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. — José Carlos Santos, Apr 24 '18 at 07:38
How many animals are there in the first year? Consider defining a recursive sequence. The set of adult cows in a particular year will be the union of the set of cows from the previous year along with the set of calves who just matured into cows who were born four years ago. Set that to mathematical symbols and apply standard known techniques. — JMoravitz, Apr 24 '18 at 07:39
@Jmoravitz, looks like the best method, is this correct: let cows in $n^{th}$ year be $C_n$ and calves be $c_n$. Then we know $C_0 = 1$ and $c_0 = 0$. Also that $C_n = C_{n-1}+c_{n-4}$ and $c_n = c_{n-1} + C_{n-1}-C_{n-4}$ — jonsno, Apr 24 '18 at 08:01
@samjoe Mmm....close, but no. Some of the calves from four years ago will have already matured into cows before this year and will have been double counted, once in $c_{n-4}$ and again in $C_{n-1}$. If you insist you could define even more sequences to use in tandem with these, one for newborn calves, one for one-year-old calves, another for two-year-old calves, etc... This technique may be more useful for problems where certain factors affect certain age-groups (such as newborn sickness, veal sales, etc...) but it is unnecessary here. You can describe the whole problem only using $C_n$. — JMoravitz, Apr 24 '18 at 08:06
@JMoravitz Hm you are correct, took some time to realise! I'll play with this for some time. — jonsno, Apr 24 '18 at 08:14

Joffan · Answer 1 · 2018-04-24T18:03:28.310

Yoou can get some insight into this problem - and indeed solve this particular case - by tracking cows and calves of different ages in a table:

\begin{array}{c|c} \text{year} & \text{mature cows} & \text{new-borns} & \text{1-y-o} & \text{2-y-o} & \text{3-y-o} & \text{total} \\ \hline 1 & 1 & 1 & 0 & 0 & 0 & 2\\ 2 & 1 & 1 & 1 & 0 & 0 & 3\\ 3 & 1 & 1 & 1 & 1 & 0 & 4\\ 4 & 1 & 1 & 1 & 1 & 1 & 5\\ 5 & 2 & 2 & 1 & 1 & 1 & 7\\ 6 & 3 & 3 & 2 & 1 & 1 & 10\\ : & : & : & : & : & : & : \\ \end{array}

After adding mmore rows, yoou should be able to see an easier way of finding the number of mmature cows (and thus new calves) in any given year.

It is perhaps indelicate to inquire into why there are no bull calves.

farruhota · Answer 2 · 2018-04-25T03:52:01.273

The computation table: $$\begin{array}{c|c|l|l|c} Years & 1 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_n&Te_n&\\ \hline 1&1_1&&&&\\ 2&1_2&&&&\\ 3&1_3&&&&\\ 4&1_4&&&&\\ 5&1_5&&&&\\ 6&1_6&1_{11}&&&\\ 7&1_7&1_{12}+1_{21}&&&\\ 8&1_8&1_{13}+1_{22}+1_{31}&&&\\ 9&1_9&1_{14}+1_{23}+1_{32}+1_{41}&&&\\ 10&1_{10}&5&&&&\\ 11&1_{11}&6&1_{111}&&\\ 12&1_{12}&7&1_{112}+1_{121}+1_{211}&&\\ 13&1_{13}&8&1_{113}+1_{122}+1_{212}+1_{131}+1_{221}+1_{311}&&\\ 14&1_{13}&9&10&&\\ 15&1_{13}&10&15&&\\ 16&1_{13}&11&21&1_{1111}&\\ 17&1_{17}&12&28&4&\\ \hline Total&18&\ \ \ \ \ \ \ \ \ \ \ \ \ \ 78&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 84&5&\color{red}{185} \end{array}$$ You can figure out the pattern for general conditions (a calf matures in $m$ years and the number of years is $n$).

score 1 · Answer 3 · answered Apr 24 '18 at 21:05

Let's indicate with :

$A(k)$ the number of adult cow at year $k$
$C_0(k)$ the number of calf born at year $k$
$C_1(k)$ the number of one-year-old calves at year $k$
$C_2(k)$ the number of two-years-old calves at year $k$
$C_3(k)$ the number of three-years-old calves at year $k$

Of course, the total population is $C(k) = A(k) + C_0(k)+C_1(k)+C_2(k)+C_3(k).$

The dynamics of the subpopulation is the following:

$$\begin{cases} A(k+1) = A(k) + C_3(k)\\ C_0(k+1) = A(k)\\ C_1(k+1) = C_0(k)\\ C_2(k+1) = C_1(k)\\ C_3(k+1) = C_2(k)\\ \end{cases}$$

By iterating this, with initial conditions $A(0) = 1$, $C_0(0) = C_1(0) = C_2(0) = C_3(0) = 0$, you get that $C(17) = 185.$

Here you can find the Matlab code (if you are familiar with it) to iterate this:

clear all
close all

A = zeros(18, 1);
C0 = zeros(18, 1);
C1 = zeros(18, 1);
C2 = zeros(18, 1);
C3 = zeros(18, 1);

A(1) = 1;

for k=1:17
    A(k+1) = A(k) + C3(k);
    C0(k+1) = A(k);
    C1(k+1) = C0(k);
    C2(k+1) = C1(k);
    C3(k+1) = C2(k);
end

C = A + C0 + C1 + C2 + C3;

figure
plot(0:17, A, 'r--', 'LineWidth', 2)
hold on
plot(0:17, C0, 'b--', 'LineWidth', 2)
plot(0:17, C1, 'g--', 'LineWidth', 2)
plot(0:17, C2, 'c--', 'LineWidth', 2)
plot(0:17, C3, 'm--', 'LineWidth', 2)
plot(0:17, C, 'k', 'LineWidth', 2)
legend('Adults', 'Brand new calves', 'One-year-old calves', 'Two-years-old calves', 'Three-years-old calves', 'Total cows')
xlim([0 17])

The (graphical) results you get is the following:

score 0 · Answer 4 · answered Apr 24 '18 at 17:00

Let $K(n)$ describe the number of cows and $C(n)$ the number of calfs in $n$-th year ($n=0,1,...$)

Notice, that:

In years $k=0,1,2,3$ we have $K(k)=1$
In $k$-th year ($k>3$) the ammount of cows increase by the number of calves born in year $k-4$, which is exactly $K(k-4)$, thus $K(k)=K(k-1)+K(k-4)$
The number of calves in $k$-th year is equal the sum of calves born in years $ k-1, k-2, k-3$, thus $C(k)=K(k-1)+K(k-2)+K(k-3)$

So:

$K(0)=K(1)=K(2)=K(3)=1$
$K(4)=2$
$K(5)=3$
$K(6)=4$
$K(7)=5$
$K(8)=7$
$K(9)=10$
$K(10)=14$
$K(11)=19$
$K(12)=26$
$K(13)=36$
$K(14)=50$
$K(15)=69$
$K(16)=95$
$K(17)=131$

And

$C(17)=214$

Number of animals is then $X=K(17)+C(17)=345$

A cow has a calf every year. A calf becomes a cow in 4 years. Starting with one cow, how many animals are there in 17 years?

4 Answers4