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Written by Adrien Barbaresi on 2024-10-23 at 18:02

I'm fiddling with Rust, using prime numbers as a playground.

A Fermi–Dirac prime is a prime power whose exponent is a power of two. Interestingly, this concept already confuses chatbots quite a bit.

I came up with the following solution which should be readable enough, can somebody tell me if there is something wrong with it?

https://oeis.org/A050376

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Written by Adrien Barbaresi on 2024-10-23 at 18:05

@HydrePrever et al. your 2¢ maybe? Interesting series related to primes?

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Written by Le Néandertal sous benzo on 2024-10-23 at 18:40

@adbar I am not sure I'm getting the algorithm actually. You first test if n is prime, but n = 4 is a Fermi-Dirac prime ; if I read correctly your function returns false in that case.

I think when n is not prime, you should look at its square root m. If m is not an integer, the case is dealt with. If m is an integer, then n is FD iff m is FD.

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Written by nours on 2024-10-24 at 04:01

@adbar I don't know enough rust (but I need to learn it), I think like in c you need to do your function with modulo (%) and not with division (/). Maybe I'm wrong.

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Written by nours on 2024-10-24 at 06:06

@adbar fn is_prime(n: u64) -> bool {

if n <= 1 {

    return false;

}

if n <= 3 {

    return true;

}

if n % 2 == 0 || n % 3 == 0 {

    return false;

}

let sqrt_n = (n as f64).sqrt() as u64;

let mut i = 5;

while i <= sqrt_n {

    if n % i == 0 || n % (i + 2) == 0 {

        return false;

    }

    i += 6;

}

true

}

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Written by nours on 2024-10-24 at 06:06

@adbar fn find_base_and_exp(n: u64) -> Option<(u64, u64)> {

// Si n est 1, ce n'est pas un primaire de Fermi-Dirac

if n == 1 {

    return None;

}

// Pour chaque nombre premier possible comme base

let sqrt_n = (n as f64).sqrt() as u64;

let mut p = 2;

while p <= sqrt_n + 1 {

    if !is_prime(p) {

        p += 1;

        continue;

    }

    let mut x = n;

    let mut exp = 0;

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Written by nours on 2024-10-24 at 06:07

@adbar // Tant que x est une puissance parfaite de p

    while x > 1 && x % p == 0 {

        x /= p;

        exp += 1;

    }

    // Si x = 1, on a trouvé notre décomposition

    if x == 1 {

        return Some((p, exp));

    }

    p += 1;

}

// Si n est lui-même premier, c'est un cas spécial (exposant 1)

if is_prime(n) {

    return Some((n, 1));

}

None

}

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Written by nours on 2024-10-24 at 06:07

@adbar fn is_power_of_two(n: u64) -> bool {

n != 0 && (n & (n - 1)) == 0

}

fn is_fermi_dirac_prime(n: u64) -> bool {

if let Some((base, exp)) = find_base_and_exp(n) {

    // Vérifie si l'exposant est une puissance de 2

    is_power_of_two(exp)

} else {

    false

}

}

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Written by Adrien Barbaresi on 2024-10-24 at 11:49

@nours Merci ! Le code pour les puissances de 2 est pratique, pour le reste je préfère passer par des bool plutôt que des Option.

J'ai le même type de calcul des nombres premiers, pas le plus rapide mais plus facile à écrire.

Je publierai le code en open source bientôt, on pourra en reparler.

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Written by nours on 2024-10-24 at 12:05

@adbar avec plaisir

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