clearvars,clc

had=zeros(1,999);

oddComp=setdiff(3:2:999,primes(999));

idx=combinator(length(oddComp),3,'c');

list=oddComp(idx);

for i=1:length(list)

  s=sum(list(i,:));

  if s<1000 

    had(s)=1;

  end

end

sum(had)

idx=combinator(length(oddComp),2,'p');

list=oddComp(idx);

for i=1:length(list)

  s=list(i,1)*2+list(i,2);

  if s<1000

    if had(s)==0

      fprintf('%d ',s);

    end

    had(s)=1;

  end

end

disp(' ')

sum(had)

 

for i=1:length(oddComp)

  s=3*oddComp(i);

  if s<1000

    if had(s)==0

      fprintf('%d ',s);

    end 

    had(s)=1;

  end

end

disp(' ')

sum(had)

finds

ans =

   476

39 33 43 53  

ans =

   480

27  

ans =

   481

If the three numbers must be unique, there are 476 such numbers; if duplicates are allowed, there are 481 such. The numbers 33, 39, 43 and 53 require that two of the numbers be the same and 27 requires all three the same (of course that's 9).