yp=char(sym(2023)^2023);

disp(length(yp))

disp(sod(yp))

finds that the 6689 digits of the power add up to 30,112:

>> yearYearn5

        6689

       30112

>> 

where the sod function is

function sd = sod(n0)

  sd=0;

  n1=char(string(n0));

  for i=1:length(n1)

   sd=sd+str2double(n1(i));

  end

end

which unnecessarily converts the character string to a unitary string and back again as the sod function is written so that it would work even on a numeric, so in fact the char function in the outer function was not really necessary except to find the length of the number, which was not really asked for in the puzzle.

So

yp=sym(2023)^2023;

disp(sod(yp))

works just as well as the above in finding only

>> yearYearn5

       30112

>>