# Irreducible polynomials over GF(5)

## Jelly, ~~30~~ ~~23~~ ~~22~~ 20 bytes

```
ÆF>1’PḄ
ÆDµU5*×Ç€S:Ṫ
```

Try it online! or verify all test cases at once.

### Algorithm

This uses the formula

$$\text{A001692}(n) = \frac 1 n \sum_{d|n} \mu(d)5^\frac n d$$

from the OEIS page, where \$d | n\$ indicates that we sum over all divisors \$d\$ of \$n\$, and \$\mu\$ represents the Möbius function.

### Code

```
ÆF>1’PḄ Monadic helper link. Argument: d
This link computes the Möbius function of d.
ÆF Factor d into prime-exponent pairs.
>1 Compare each prime and exponent with 1. Returns 1 or 0.
’ Decrement each Boolean, resulting in 0 or -1.
P Take the product of all Booleans, for both primes and exponents.
Ḅ Convert from base 2 to integer. This is a sneaky way to map [0, b] to
b and [] to 0.
ÆDµU5*×Ç€S:Ṫ Main link. Input: n
ÆD Compute all divisors of n.
µ Begin a new, monadic chain. Argument: divisors of n
U Reverse the divisors, effectively computing n/d for each divisor d.
5* Compute 5 ** (n/d) for each n/d.
Ç€ Map the helper link over the (ascending) divisors.
× Multiply the powers by the results from Ç.
S Add the resulting products.
:Ṫ Divide the sum by the last divisor (n).
```

## Mathematica, ~~39~~ 38 bytes

```
DivisorSum[a=#,5^(a/#)MoebiusMu@#/a&]&
```

Uses the same formula as the Jelly answer.