reserve a,b,c,k,m,n for Nat;
reserve i,j for Integer;
reserve p for Prime;

theorem
  for n being non zero Nat holds
  n > 1 implies 1/3 * (2|^(2|^(n+1)) + 2|^(2|^n) + 1) is composite
  proof
    let n be non zero Nat;
    assume
A1: n > 1;
    set m = 2|^(2|^(n+1)) + 2|^(2|^n) + 1;
    1/3*m >= 1/3*7 by Th39,XREAL_1:64;
    hence 1/3*m >= 2 by XXREAL_0:2;
A2: 7 divides m by Th45;
    3 divides m by Th44;
    then 7*3 divides m by A2,PEPIN:4,INT_2:30,XPRIMES1:3,7;
    then consider k such that
A3: m = 7*3*k;
    now
      assume k < 2;
      then per cases by NAT_1:23;
      suppose k = 0;
        hence contradiction by A3;
      end;
      suppose k = 1;
        hence contradiction by A1,A3,Th41;
      end;
    end;
    then 7*k is composite by NUMBER10:3;
    hence thesis by A3;
  end;
