Centerless Polygons

Haskell, 42 bytes

f n=sum[0^mod n q|a<-[3..n],q<-[a,3*a..n]]

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51 bytes

f n=sum[1|a<-[3..n],b<-[1,3..n],c<-[1..n],a*b*c==n]

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The output is the number of ways to factor \$n=abc\$ into three positive factors, where \$a \geq 3\$, \$b\$ is odd, and \$c\$ is unconstrained.

05AB1E, 17 16 10 bytes


Try it online! Edit: Saved 5 bytes thanks to @ovs. Explanation:

3÷L         Get a list from `0` to `n//3`.
   sÑÃ      Keep only factors of `n`.
      ÑÉO   Number of odd divisors of each factor.
         O  Output the sum.

J, 29 bytes

[:+/@,[=[+/\\[email protected](*1+i.)~"+3+i.

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[:+/@,[=[+/\\[email protected](*1+i.)~"+3+i.
                           i. 0…N-1
                         3+   3…N+2
                       "+     for each y in 3…N+2:
        [      (*1+i.)~       y * 0…N, thus f.e. 5 10 15 20 … for p_5
           \\[email protected]               take every possible sublist
         +/                   and sum it
      [=                      which sums are equal to N?
[:+/@,                        count the true bits