/* The Liouville function on a integer n is L(n) = (-1)^r where r is the number of prime factors in the prime factorization of n (L(1) = 1). This program is called as sumliouville begin end and prints the sum of L(n) for n from 'begin' to 'end'. Compile with gcc -O2 -Wall -g -o sumliouville sumliouville.c -lm
This program is an interesting example with respect to starting several of them at the same time. On an Intel quad-core with hyperthreading I found that the single processes can become a factor of 6 and more slower when 8 instances of this program are started at the same time.
The slightly more complicated variant sumliouville2 is more efficient for well chosen interval lengths, and behaves well when several instances are run in parallel.
*/
inlinevoid do_p(long p)
{ long q, k, start; for (q = p; q <= end; q *= p){
start = (-begin) % q + 1; if (start <= 0) start += q; for (k = start; k <= l; k += q){
res[k] = -res[k];
found[k] *= p;
}
} /* printf("\n"); for (k=1; k<=l; k++) printf("%ld ",(long)res[k]); printf("\n"); for (k=1; k<=l; k++) printf("%ld ",found[k]);
printf("\n"); */
}
int main(int argc, char *argv[])
{ long i, k, p, n, n2, off, sum; char *erat;
begin = atol(argv[1]);
end = atol(argv[2]);
/* compute primes */
n = lsqrt(end)+1;
n2 = n/2;
erat = (char*) malloc((n2+1) * sizeof(char)); for (i=0; (2*i+1)*(2*i+1) <= n; ) {
i++; while (erat[i] == 1) i++;
p = 2*i+1; for (k = (p*p-1)/2; k <= n2; k += p) erat[k] = 1;
}
/* init memory */
l = end - begin + 1;
res = (char*) malloc((l+1)*sizeof(char)); for(i=1; i<=l; i++) res[i] = 1;
found = (long*) malloc((l+1)*sizeof(long)); for(i=1; i<=l; i++) found[i] = 1;
/* mark parity of number of prime factors */
do_p(2); for (i = 1; i <= n2; i++) { if (erat[i] == 0) {do_p(2*i + 1); }//printf("%ld ",2*i+1); fflush(stdout);}
}
/* compute sum */
sum = 0;
off = begin-1; for (i=1; i<=l; i++) { if (found[i] < off+i) sum -= res[i]; /* one more prime */ else sum += res[i];
}
printf("%ld\n", sum); return 0;
}
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