As we start with a number n which is either 1 or 2 (mod 3) and 0 (mod 2), it means this is either 2 or 4 (mod 6). If its sigma is 0 (mod 3), then we have the next term t=2*sigma-n is either t=2*0-1=2 (mod 3) or t=2*0-2=1 (mod 3). So the only combinations that would result in 0 (mod 3) are either (A) n=4 (mod 6) and sigma(n)=2 (mod 3) or (B) n=2 (mod 6) and sigma(n)=1 (mod 3). We can expand some

formulae for sigma and see if we can get this, considering that if n=1 (mod 3) than its prime factors which are 2 (mod 3) can be grouped two-by-two in those products for sigma, etc.