There is an operation program that can make (a + 1) ⊕ B = n + 1, a ⊕ B + 1 = n-2 when a ⊕ B = n (n is a constant). Now we know that 1 ⊕ 1 = 2, then 3 ⊕ 3=______ ．

There is an operation program that can make (a + 1) ⊕ B = n + 1, a ⊕ B + 1 = n-2 when a ⊕ B = n (n is a constant). Now we know that 1 ⊕ 1 = 2, then 3 ⊕ 3=______ ．

Now we know that 1 ⊕ 1 = 2, finding 3 ⊕ 3 is equivalent to a increasing 2, B increasing 2, the result is to increase 2 and decrease 4 on the basis of 2, that is, 2 + 2-4 = 0

There is an operation program, can make a ⊕ B = n (n is a constant), get (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2=______ ．

⊙ 1 ⊕ 1 = 2 (where a = 1, B = 1, n = 2) ⊕ 2 ⊕ 1 = 3, 2 ⊕ 2 = 1 (where a = 2, B = 2, n = 1), 3 ⊕ 2 = 2, 3 ⊕ 3 = 0 (where a = 3, B = 3, n = 0) ⊕ 4 ⊕ 3 = 14 ⊕ 4 = - 15 ⊕ 5 = - 2 Therefore, the answer is - 2006

Ask smart people to help! There is an operation program that can make: a @ b = n (n is a constant)
When (a + 1) @ b = n + 1, a @ (B + 1) = n + 2, then (a + 2) @ (B + 2)=______ (expressed by the algebraic expression of n) @ stands for a special sign with no special meaning

Tip: add one to the left and add one to the result; add one to the right and add two to the result