Removal of For Loops

Is it possible to script the following function in MATLAB without a for loop, or any other iterative loop? I am trying to understand if it is possible to have a matrix reference itself as it is populated using a single command in MATLAB. Thanks for looking!

x(1)=0;
x(2)=1;
x(3)=2;

for n=4:51
    x(n)=x(n-1) + x(n-3);
end

Answer 1

[Sean de Wolski]

It's probably possible with filter() or similar. But I guarantee a well written for-loop will not be much slower. Just remember to preallocate x.

x = zeros(51,1); 
%etc.

Answer 2

[Walter Roberson]

I = sqrt(-1);
x = @(n) -(301/14945472)*((1+I*3^(1/2))*(93^(1/2)-93/7)*(108+12*93^(1/2))^(1/3)-744/7-(5/14)*(-1+I*3^(1/2))*(93^(1/2)-31/5)*(108+12*93^(1/2))^(2/3))*(((1/72)*(-93^(1/2)+9)*(108+12*93^(1/2))^(2/3)+(1/6)*(108+12*93^(1/2))^(1/3))^n*(((I*3^(1/2)-67/43)*93^(1/2)+279/43-((775/43)*I)*3^(1/2))*(108+12*93^(1/2))^(1/3)+((-((97/129)*I)*3^(1/2)+27/43)*93^(1/2)+((248/43)*I)*3^(1/2)-310/43)*(108+12*93^(1/2))^(2/3)+1860/43+((92/43)*I)*3^(1/2)*93^(1/2))*((1/72)*(-9+93^(1/2))*(-1+I*3^(1/2))*(108+12*93^(1/2))^(1/3)-(1/72)*(108+12*93^(1/2))^(2/3)-((1/72)*I)*(108+12*93^(1/2))^(2/3)*3^(1/2)+1/3)^n+(6696/43)*(-(1/72)*(-9+93^(1/2))*(1+I*3^(1/2))*(108+12*93^(1/2))^(1/3)-(1/72)*(108+12*93^(1/2))^(2/3)+((1/72)*I)*(108+12*93^(1/2))^(2/3)*3^(1/2)+1/3)^n*((1/72)*(-93^(1/2)+9)*(108+12*93^(1/2))^(2/3)+(1/6)*(108+12*93^(1/2))^(1/3))^n+((-((12/43)*I)*3^(1/2)+98/43)*93^(1/2)-1302/43-((248/43)*I)*3^(1/2))*(108+12*93^(1/2))^(1/3)+((-((89/129)*I)*3^(1/2)+35/43)*93^(1/2)+((279/43)*I)*3^(1/2)-217/43)*(108+12*93^(1/2))^(2/3)+1860/43-((92/43)*I)*3^(1/2)*93^(1/2))/((1/72)*(-93^(1/2)+9)*(108+12*93^(1/2))^(2/3)+(1/6)*(108+12*93^(1/2))^(1/3))^n

But watch out for floating point round-off.

(Yes, really. And yes, this is the simplified form of the expression.)

[Sean de Wolski]

nice! +1 for sure.

[Walter Roberson]

Note: the above was produced by simplifying the output of Maple's rsolve() routine. The basic form of the answer is not very complicated, but it involves the sum of terms with the sum taken over the roots of a cubic expression, and that cubic happens to have two imaginary solutions. The expanded expression before simplification is pretty grotty.