What is the number of possible relationships between N numbers?

Question

Mathematics

Obviously, for two numbers, this number is 3:

1. A=B
   
2. A < B
   
3. B<A
   

In the case of three numbers, we find that there are 13 options:

1. A=B=C
   
2. A<B<C
   
3. B<A<C
   
4. A<C<B
   
5. C<A<B
   
6. B<C<A
   
7. C<B<A
   
8. A=B<C
   
9. C<A=B
   
10. B=C<A
    
11. A<B=C
    
12. A=C<B
    
13. B<A=C
    

We argue as follows: when all three numbers are equal, this is one option; when all three numbers are not equal to each other - that's 3 more!=6 options; when two of some numbers are equal to each other and not equal to the third - this is 3 more (the number of possible combinations of 3 numbers by 2) multiplied by 2 (more or less than the third number) = 6 options.
Arguing in this way, we can calculate that for 4 numbers there are 81 options, and for five - 651. But the enumeration of cases becomes more and more complex and confusing. Is there a formula that reflects the dependence of the number of relations between N numbers on the number of these numbers?
PS. Search in the encyclopedia of numerical sequences did not return results.

Answer 1

In general, your sequence is similar to https://oeis.org/A000670:
Number of ways n competitors can rank in a competition, allowing for the possibility of ties.
Also number of asymmetric generalized weak orders on n points.
Also called the ordered Bell numbers.

Answer 2

If we take only letter designations, without being tied to a number, then the formula is as follows: the
number of permutations of the LETTERS * the number of permutations of the used characters of the
letter: A, B
characters: =, <, >
the number of permutations of the letters: AB, BA - 2
the number of permutations of the characters: =, < , > - 3
2 * 3 = 6
1) A = B
2) B = A
3) A < B
4) B < A
5) A > B
6) B > A