# 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

```
3÷ÝsÑÃÑÉOO
```

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
```