Brilliant Butterflies

Here's my simple solution: Think about the gametes produced.

During each generation, the amounts of each gametes, R and b, are conservated and their ratio is always the same .

When the red butterflies breed with each other, according to the Punnett square, the heterozygote (Rb) will gradually turn into RR, and bb, therefore the portion of the heterozygote will be approaching to $0$ , only left with RR and bb.

Since the initial portion is $\dfrac{1}{4}$ RR, $\dfrac{1}{4}$ Rb, $\dfrac{1}{2}$ bb, the portion of the gametes, R and b, is $\dfrac{3}{8}$ and $\dfrac{5}{8}$ .

And since the the amounts of each gametes are conservated , the final state is $\dfrac{3}{8}$ RR and $\dfrac{5}{8}$ bb,

Thus the portion of blue butterflies is $\boxed{\dfrac{5}{8}}$ , $a+b=\boxed{13}$ .

Let the proportions of large red, small red and blue butterflies after $n$ generations be $R_n,r_n,b_n$ , respectively. Then $R_n + r_n + b_n = 1$ for all $n$ , while $\begin{aligned} R_{n+1} & = \; R_n\left(\lambda_n + \tfrac12(1-\lambda_n)\right) + r_n\left(\tfrac12\lambda_n + \tfrac14(1-\lambda_n)\right) \; = \; \tfrac12(1 + \lambda_n)R_n + \tfrac14(1 + \lambda_n)r_n \\ r_{n+1} & = \; R_n \tfrac12(1 - \lambda_n) + r_n\left(\tfrac12\lambda_n + \tfrac12(1-\lambda_n)\right) \; =\; \tfrac12(1-\lambda_n)R_n + \tfrac12r_n \end{aligned}$ where $\lambda_n = \frac{R_n}{R_n + r_n}$ is the proportion of large red butterflies in the red butterfly population. Thus $\begin{aligned} R_{n+1} & = \; \frac{(2R_n + r_n)^2}{4(R_n + r_n)} \\ r_{n+1} & = \; \frac{r_n(2R_n + r_n)}{2(R_n + r_n)} \end{aligned}$ From these identities it follows that $2R_{n +1} + r_{n+1} = 2R_n + r_n$ for all $n \ge 0$ , and hence that $2R_n + r_n = 2R_0 + r_0$ for all $n \ge 0$ . This implies that $\begin{aligned} r_{n+1} & = \; (2R_0 + r_0) \frac{r_n}{2R_0 + r_0 + r_n} \\ r_{n+1}^{-1} & = \; r_n^{-1} + (2R_0 + r_0)^{-1} \end{aligned}$ so that $\begin{aligned} r_n^{-1} & = \; r_0^{-1} + n(2R_0 + r_0)^{-1} \;=\; \frac{2R_0 + r_0 + nr_0}{r_0(2R_0+r_0)} \\ r_n & = \; \frac{r_0(2R_0 + r_0)}{2R_0 + r_0 + nr_0} \end{aligned}$ Thus we can find exact expressions for $R_n$ , $r_n$ and $b_n$ . Letting $n \to \infty$ , we deduce that $\begin{aligned} \lim_{n \to \infty} r_n & = \; 0 \\ \lim_{n \to \infty}R_n & = \; R_0 + \tfrac12r_0 \\ \lim_{n \to \infty}b_n & = \; 1 - 0 - (R_0+ \tfrac12r_0) \; = \; \tfrac12r_0 + b_0 \end{aligned}$ With $r_0 = \tfrac14$ and $b_0 = \tfrac12$ , we obtain $\lim_{n \to \infty} b_n = \tfrac58$ , making the answer $\boxed{13}$ .

Let the proportions of large red, small red and blue butterflies in the $n^{\text{th}}$ generation be $R_n,r_n,b_n$ .

Two large red butterflies mate with probability $\frac{R_n^2}{R_n+r_n}$ , and their offspring will be a large red butterfly with probability $1$ .

A large red mates with a small red with probability $\frac{2R_n r_n}{R_n+r_n}$ , and their offspring will be a large red butterfly with probability $\frac12$ .

Two small reds mate with probability $\frac{r_n^2}{R_n+r_n}$ , and their offspring will be a large red butterfly with probability $\frac14$ .

Continuing this analysis, we get the proportions in the next generation $\begin{aligned} R_{n+1}&=\frac{(2R_n+r_n)^2}{4(R_n+r_n)} \\ r_{n+1}&=\frac{r_n (2R_n+r_n)}{2(R_n+r_n)} \\ b_{n+1}&=b_n+\frac{r_n^2}{4(R_n+r_n)} \end{aligned}$

First, we will prove that the proportion of small reds tends to zero as $n \to \infty$ . Assume that $r_1>0$ (as it is in this problem), and consider the ratio $a_n=\frac{R_n}{r_n}$ . We have $\begin{aligned} a_{n+1}&=\frac{2R_n+r_n}{2r_n} \\ &=\frac{R_n}{r_n}+\frac12 \\ &=a_n+\frac12 \end{aligned}$

So clearly $a_n \to \infty$ . Since $R_n$ is certainly finite, the only way this can happen is if $r_n \to 0$ , as required. Hence $b_n+R_n \to 1$ .

Now note that the quantity $b_n-R_n$ does not change from one generation to the next: $\begin{aligned} b_{n+1}-R_{n+1} &=b_n+\frac{r_n^2}{4(R_n+r_n)} - \frac{(2R_n+r_n)^2}{4(R_n+r_n)} \\ &=b_n+\frac{r_n^2-(2R_n+r_n)^2}{4(R_n+r_n)} \\ &=b_n-\frac{4R_n r_n + R_n^2}{4(R_n+r_n)} \\ &=b_n-R_n \end{aligned}$

With these facts, we can solve the problem; we have $b_n+R_n \to 1$ as $n \to \infty$ and $b_n-R_n=b_0-R_0=\frac14$ for all $n$ .

Hence $b_n \to \frac58$ as $n \to \infty$ and the required answer is $5+8=\boxed{13}$ .

The Punnett square probabilities are:

An RR with RR will always make RR, an RR with Rb will make $\frac{1}{2}$ RR and $\frac{1}{2}$ Rb, an Rb with Rb will make $\frac{1}{4}$ RR, $\frac{1}{2}$ Rb, and $\frac{1}{4}$ bb, and a bb with bb always make bb.

Let's say the ratio of the number of large red butterflies (RR) to the whole population is $x$ , the ratio of small red butterflies (Rb) to the whole population is $y$ , and the ratio of blue butterflies (bb) to the whole population is $z$ .

Then $\frac{x}{x + y}$ of the large red butterflies (RR) will breed with large red butterflies (RR), and $\frac{y}{x + y}$ of the large red butterflies (RR) will breed with small red butterflies (Rb). Similarly, $\frac{x}{x + y}$ of the small red butterflies (Rb) will breed with large red butterflies (RR), and $\frac{y}{x + y}$ of the small red butterflies (Rb) will breed with small red butterflies (Rb). All blue butterflies (bb) will breed with other blue butterflies (bb). This means RR will breed with RR $\frac{x^2}{x + y}$ of the time, RR will breed with Rb $\frac{xy}{x + y} + \frac{xy}{x + y} = \frac{2xy}{x + y}$ of the time, Rb will breed with Rb $\frac{y^2}{x + y}$ of the time, and bb will breed with bb $z$ of the time.

RR with RR: $\frac{x^2}{x + y}$

RR with Rb: $\frac{2xy}{x + y}$

Rb with Rb: $\frac{y^2}{x + y}$

bb with bb: $z$

Using the Punnett Square probabilities above, an RR will always be bred from RR and RR parents, $\frac{1}{2}$ of the time from RR and Rb parents, and $\frac{1}{4}$ of the time from Rb and Rb parents, for a total of $\frac{x^2}{x + y} + \frac{1}{2} \cdot \frac{2xy}{x + y} + \frac{1}{4} \cdot \frac{y^2}{x + y} = \frac{4x^2 + 4xy + y^2}{4(x + y)}$ of the population.

RR from:

RR with RR: $1 \cdot \frac{x^2}{x + y}$

RR with Rb: $\frac{1}{2} \cdot \frac{2xy}{x + y}$

Rb with Rb: $\frac{1}{4} \cdot \frac{y^2}{x + y}$

Total: $\frac{4x^2 + 4xy + y^2}{4(x + y)}$

Likewise, an Rb will be bred $\frac{1}{2}$ of the time from RR and Rb parents, and $\frac{1}{2}$ of the time from Rb and Rb parents, for a total of $\frac{1}{2} \cdot \frac{2xy}{x + y} + \frac{1}{2} \cdot \frac{y^2}{x + y} = \frac{2xy + y^2}{2(x + y)}$ of the population.

Rb from:

RR with Rb: $\frac{1}{2} \cdot \frac{2xy}{x + y}$

Rb with Rb: $\frac{1}{2} \cdot \frac{y^2}{x + y}$

Total: $\frac{2xy + y^2}{2(x + y)}$

Finally, a bb will be bred $\frac{1}{4}$ of the time from Rb and Rb parents, and always from bb and bb parents, for a total of $\frac{1}{4} \cdot \frac{y^2}{x + y} + z = \frac{y^2}{4(x + y)} + z$ of the population.

bb from:

Rb with Rb: $\frac{1}{2} \cdot \frac{y^2}{x + y}$

bb with bb: $1 \cdot z$

Total: $\frac{y^2}{4(x + y)} + z$

There is therefore a recursive relationship of:

$x_{n + 1} = \frac{4x_n^2 + 4x_ny_n + y_n^2}{4(x_n + y_n)}$

$y_{n + 1} = \frac{2x_ny_n + y_n^2}{2(x_n + y_n)}$

$z_{n + 1} = \frac{y_n^2}{4(x_n + y_n)} + z_n$

for each generation $n$ , where $x_0 = \frac{1}{4}$ , $y_0 = \frac{1}{4}$ , and $z_0 = \frac{1}{2}$ .

This recursive relationship can be shown inductively to have the general equations of:

$x_n = \frac{3n + 6}{8n + 24}$

$y_n = \frac{3}{4n + 12}$

$z_n = \frac{5n + 12}{8n + 24}$

.

Inductive Proof:

The first case is true because $x_0 = \frac{3(0) + 6}{8(0) + 24} = \frac{1}{4}$ , $y_0 = \frac{3}{4(0) + 12} = \frac{1}{4}$ , and $z_0 = \frac{5(0) + 12}{8(0) + 24} = \frac{1}{2}$ . Assuming $x_n = \frac{3n + 6}{8n + 24}$ , $y_n = \frac{3}{4n + 12}$ , and $z_n = \frac{5n + 12}{8n + 24}$ are true:

$x_{n + 1} = \frac{4x_n^2 + 4x_ny_n + y_n^2}{4(x_n + y_n)} = \frac{4(\frac{3n + 6}{8n + 24})^2 + 4(\frac{3n + 6}{8n + 24})(\frac{3}{4n + 12}) + (\frac{3}{4n + 12})^2}{4(\frac{3n + 6}{8n + 24} + \frac{3}{4n + 12})} = \frac{3(n + 1) + 6}{8(n + 1) + 24}$

$y_{n + 1} = \frac{2x_ny_n + y_n^2}{2(x_n + y_n)} = \frac{2(\frac{3n + 6}{8n + 24})(\frac{3}{4n + 12}) + (\frac{3}{4n + 12})^2}{2(\frac{3n + 6}{8n + 24} + \frac{3}{4n + 12})} = \frac{3}{4(n + 1) + 12}$

$z_{n + 1} = \frac{y_n^2}{4(x_n + y_n)} + z_n = \frac{(\frac{3}{4n + 12})^2}{4(\frac{3n + 6}{8n + 24} + \frac{3}{4n + 12})} + \frac{5n + 12}{8n + 24} = \frac{5(n + 1) + 12}{8(n + 1) + 24}$

which completes the inductive proof.

As time goes on, the expected portion of blue butterflies will be:

$\displaystyle \lim_{n\to\infty} z_n$

$= \displaystyle \lim_{n\to\infty} \frac{5n + 12}{8n + 24}$

$= \displaystyle \lim_{n\to\infty} \frac{5 + \frac{12}{n}}{8 + \frac{24}{n}}$

$= \displaystyle \frac{5}{8}$

so $a = 5$ , $b = 8$ , and $a + b = \boxed{13}$ .