Let $a =\lim a_n$ and $x_n = 3a_{n+2} + 2a_{n+1} + a_n$ So, $x_{n+1} = x_n$ . Indeed, $3a_{n+3} = a_{n+2} + a_{n+1} + a_n \Leftrightarrow$ $3a_{n+3} + 2a_{n+2} + a_{n+1} = 3a_{n+2} + 2a_{n+1} + a_n \Leftrightarrow$ $x_{n+1}=x_n$ Then, $x_n = x_1 = 3^2 + 2^2+1^2= 14$ . In other hand, we have $\lim x_n = \lim 3a_{n+2} + 2a_{n+1} + a_n = 3a + 2a + a = 6a \Rightarrow$ $a = \dfrac{\lim x_n}{6} = \dfrac{x_1}{6} = \dfrac{7}{3}$

$3a_4 = a_1 + a_2 + a_3$ $3a_5 = a_2 + a_3 + a_4$ $3a_6 = a_3 + a_4 + a_5$ $3a_7 = a_4 + a_5 + a_6$ and so on and on finally converging to $3a_@ = a_@ + a_@ + a_@$ $3a_@ = a_@ + a_@ + a_@$ The subscript @ has been used to show the term at infinity (converged). Now adding all these equations three $a_4 s$ in L.H.S. cancel with three $a_4 s$ in R.H.S. and so do all others. We finally get $6a_@ = a_1 + 2a_2 + 3a_3$ Now it is easy to see that the rational number is (1 + 2x2 + 3x3)/6 = 7/3. Hence a = 7, b = 3. a + b = 10

Weird solution:

The recursive operation is given by the matrix:

$A=\left(\begin{array}{c c c} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array}\right)$

We want to find:

$\lim_{n\rightarrow\infty}A^n\left(\begin{array}{c}3 \\ 2 \\ 1 \\ \end{array}\right)$

which we can do by diagonalizing A. After some pain you end up with: $\left(\begin{array}{c c c} \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ ? & ? & ? \\ ? & ? & ? \\ \end{array}\right) \left(\begin{array}{c} 3 \\ 2 \\ 1 \\ \end{array}\right)= \left(\begin{array}{c} \frac{7}{3} \\ \frac{7}{3} \\ \frac{7}{3} \\ \end{array}\right)$

where we don't care what the ?'s are because we know that as the sequence converges our column vector elements become equivalent. Hence our final result is $\frac{7}{3}$ and $a+b=\boxed{10}$

include <iostream>

using namespace std;

int main() { // your code goes here float a[10001]; a[0]=1; a[1]=2; a[2]=3; for(int i=3;i<10001;i++){ a[i] =a[i-1]+a[i-2]+a[i-3]; a[i]/=3; } cout<<a[10000]; }