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- #!/usr/bin/perl
- # Daniel "Trizen" Șuteu
- # Date: 20 July 2020
- # https://github.com/trizen
- # Smallest prime factor
- # https://projecteuler.net/problem=521
- # For each prime p < sqrt(n), we count how many integers k <= n have lpf(k) = p.
- # We have G(n,p) = number of integers k <= n such that lpf(k) = p.
- # G(n,p) can be evaluated recursively over primes q < p.
- # Equivalently, G(n,p) is the number of p-rough numbers <= floor(n/p);
- # There are t = floor(n/p) integers <= n that are divisible by p.
- # From t we subtract the number integers that are divisible by smaller primes than p.
- # The sum of the primes is p * G(n,p).
- # When G(n,p) = 1, then G(n,p+r) = 1 for all r >= 1.
- # Runtime: 2.5 seconds (when Kim Walisch's `primesum` tool is installed).
- use 5.020;
- use integer;
- use ntheory qw(:all);
- use Math::Sidef qw();
- use experimental qw(signatures);
- local $Sidef::Types::Number::Number::USE_PRIMESUM = 1;
- my $MOD = 1e9;
- sub S($n) {
- my $sum = 0;
- my $s = sqrtint($n);
- forprimes {
- $sum += mulmod($_, rough_count($n/$_, $_), $MOD);
- } $s;
- addmod($sum, Math::Sidef::sum_primes(next_prime($s), $n) % $MOD, $MOD);
- }
- say S(1e12);
- __END__
- S(10^1) = 28
- S(10^2) = 1257
- S(10^3) = 79189
- S(10^4) = 5786451
- S(10^5) = 455298741
- S(10^6) = 37568404989
- S(10^7) = 3203714961609
- S(10^8) = 279218813374515
- S(10^9) = 24739731010688477
- S(10^10) = 2220827932427240957
- S(10^11) = 201467219561892846337
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