Let A1, A2, A3 ··· an be an arrangement of 1,2,3 ··· n, and n be an odd number, then (A1-1) (A2-2) (A3-3) ··· (an-n) is an odd number Even? Why

Let A1, A2, A3 ··· an be an arrangement of 1,2,3 ··· n, and n be an odd number, then (A1-1) (A2-2) (A3-3) ··· (an-n) is an odd number Even? Why

Because n is an odd number, there are odd numbers here, and the number of odd numbers is 1 more than that of even numbers. We know that odd numbers - odd numbers = even numbers, even numbers - even numbers = even numbers. No matter how 1,2,3.. n is arranged, at least one ai-i = even number will appear. (you can try n = 1-7 for example). We also know that even numbers multiplied by any number are even numbers, so the result must be even numbers

Let A1, A2 An is 1, 2 1 / 2 + 2 / 3 +... + (n-1) / N ≤ A1 / A2 + A2 / A3 +... + an-1 / an

=It's in order
Any pair of inversions will get the result that the extreme condition is all inversions, that is, the reverse order
On the basis of sequence, any an-1 / an + am-1 / am - (an-1 / am + am-1 / an) > 0 if n

Let the reverse order number of n-ary permutation A1 A2 A3 ` an be K, then what is an '` A3 A2 A1
Just be more specific

(the inverse number of A1, A2... An) + (the inverse number of an... A2, A1) = fixed value
How to find the fixed value?
If the permutation is arranged from small to large, the inverse ordinal number is 0;
And then reverse the permutation to get a decreasing permutation from large to small,
Its reverse order number is (n-1) + (n-2) +... + 2 + 1 = (n-1) n / 2,
The fixed value is (n-1) n / 2
Then the result is (n-1) n / 2-k