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.

Answer 1

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.

Answer 2

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 $n$).

Answer 3

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 $X=K(17)+C(17)=345$

Answer 4

Let's indicate with :

1. $A(k)$ the number of adult cow at year $k$
2. $C_0(k)$ the number of calf born at year $k$
3. $C_1(k)$ the number of one-year-old calves at year $k$
4. $C_2(k)$ the number of two-years-old calves at year $k$
5. $C_3(k)$ 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 $C(17) = 185.$

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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: