https://pastebin.com/dMiZcNLA
can someone help me with this?
Not the one related to june contest
Why don’t you paste the problem here?
The first obvious thing to notice is that the permutations of divisors can be replaced by permutations of the sequence \{1, 2, 3 \ldots d\} where d is the number of divisors of n.
The problem then becomes: Find the number of permutations p of \{1 \ldots n\} such that p_{i+1} \neq p_i + 1 for all valid i.
I just wrote a brute force in python for the first few terms and putting that into oeis.org gives A000255. I have provided what I believe is a step in the right direction towards the proof but it is incomplete. Sorry if you were looking for a proof and not just a solution ( you can probably still find proofs for the formula there ).
You can calculate the number of divisors of all numbers from 1 to 120000 using a modified sieve.
Attempt at proof:
Define an unwanted position in a permutation p be a position i such that p_{i+1} = p_i + 1
Consider a permutation of \{ 1, 2 \ldots n\} which has k unwanted positions . Now you can insert the element n+1 at n+1 places to create a valid permutation of \{1, 2, \ldots ,n+1 \} in the following ways:
 You put it between any one of the unwanted positions and get a permutation with k1 unwanted positions ( Note that n can’t be at an unwanted position ). You can do this in k ways.
 You put it just after n and get a permutation with k+1 unwanted positions.
 You put it at any remaining position and get a permutation with k unwanted positions. You can do this in nk ways.
If we let T(n,k) be the number of permutations of \{1,2\ldots ,n\} with k unwanted positions and combine the above we get the following recurrence T(n,k) = (k+1)\cdot T(n1, k+1) + (n1k)\cdot T(n1, k) + T(n1,k1).
I have verified this recurrence. SOURCE
 Mr. X is teaching number theory in his class. He is discussing about factors and permutations in his class. A factor of a positive integer N is a positive integer that divides n precisely (without leaving a remainder). The set of factors always includes 1 and N.

Mr. X likes combinatorics a lot. He asked his students find out all the factors of the number Y, and sort them in an ascending order. He asks them to list all permutations of the factors. They then need to cross out all permutations where two adjacent numbers are adjacent in the same order in the original list. The number of uncrossed (valid) permutations are to be given to him.

Illustration:

Integer 9 has 3 factors [1,3,9].

The permutations of these factors of number 9 are [1,3,9],[1,9,3],[3,9,1],[3,1,9],[9,1,3],[9,3,1].

Of these 6 permutations, we need to cross out [1,3,9] (1 3 adjacent in same order), [3,9,1] (3 9 in same order) and[9,1,3] (1 3 in same order)

The remaining (valid) permutations are:

[1,9,3] ,[9,3,1],[3,1,9]

Hence the number of valid permutations =3, which is the answer.

Constraints
 1<= N<=120000

1<=T <= 100

Input Format
 The first line contains T, the number of testcases

Next T lines contain the integer N

Output
 T lines containing number of valid permutations satisfying conditions mentioned in the problem statement for given input.

Test Case

Explanation
 Example 1

Input

1

10

Output

11

Explanation

T=1 (there is 1 test case)

N=10.

10 has 4 factors [1,2,5,10]. There are 24 permutations of these four factors. The 11 valid permutations are [1,5,2,10],[1,10,5,2], [2,1,10,5],[2,10,1,5], [2,10,5,1], [5,1,10,2], [5,2,1,10], [5,2,10,1], [10,1,5,2], [10,2,1,5], [10,5,2,1]. Hence the output is 11

Example 2

Input

2

6

9

Output

11

3

Explanation

T=2 (there are 2 test cases).

In the first test case, N=6. 6 has four factors [1,2,3,6]. As in the previous example, there are 11 valid permutations for these. Hence the output for the first test case is 11. This is the first line of the output

In the second test case, N=9. As was shown in the Illustration in the problem statement, thenumber of valid permutations is 3. Hence the output for the second test case (the second line of the output) is 3.