Task: Mertens function

C: #include <stdio.h>
#include <stdlib.h>

int* mertens_numbers(int max) {
    int* m = malloc((max + 1) * sizeof(int));
    if (m == NULL)
        return m;
    m[1] = 1;
    for (int n = 2; n <= max; ++n) {
        m[n] = 1;
        for (int k = 2; k <= n; ++k)
            m[n] -= m[n/k];
    }
    return m;
}

int main() {
    const int max = 1000;
    int* mertens = mertens_numbers(max);
    if (mertens == NULL) {
        fprintf(stderr, "Out of memory\n");
        return 1;
    }
    printf("First 199 Mertens numbers:\n");
    const int count = 200;
    for (int i = 0, column = 0; i < count; ++i) {
        if (column > 0)
            printf(" ");
        if (i == 0)
            printf("  ");
        else
            printf("%2d", mertens[i]);
        ++column;
        if (column == 20) {
            printf("\n");
            column = 0;
        }
    }
    int zero = 0, cross = 0, previous = 0;
    for (int i = 1; i <= max; ++i) {
        int m = mertens[i];
        if (m == 0) {
            ++zero;
            if (previous != 0)
                ++cross;
        }
        previous = m;
    }
    free(mertens);
    printf("M(n) is zero %d times for 1 <= n <= %d.\n", zero, max);
    printf("M(n) crosses zero %d times for 1 <= n <= %d.\n", cross, max);
    return 0;
}





Java: public class MertensFunction {

    public static void main(String[] args) {
        System.out.printf("First 199 terms of the merten function are as follows:%n    ");
        for ( int n = 1 ; n < 200 ; n++ ) {
            System.out.printf("%2d  ", mertenFunction(n));
            if ( (n+1) % 20 == 0 ) {
                System.out.printf("%n");
            }
        }
        
        for ( int exponent = 3 ; exponent<= 8 ; exponent++ ) {
            int zeroCount = 0;
            int zeroCrossingCount = 0;
            int positiveCount = 0;
            int negativeCount = 0;
            int mSum = 0;
            int mMin = Integer.MAX_VALUE;
            int mMinIndex = 0;
            int mMax = Integer.MIN_VALUE;
            int mMaxIndex = 0;
            int nMax = (int) Math.pow(10, exponent);
            for ( int n = 1 ; n <= nMax ; n++ ) {
                int m = mertenFunction(n);
                mSum += m;
                if ( m < mMin ) {
                    mMin = m;
                    mMinIndex = n;
                }
                if ( m > mMax ) {
                    mMax = m;
                    mMaxIndex = n;
                }
                if ( m > 0 ) {
                    positiveCount++;
                }
                if ( m < 0 ) {
                    negativeCount++;
                }
                if ( m == 0 ) {
                    zeroCount++;
                }
                if ( m == 0 && mertenFunction(n - 1) != 0 ) {
                    zeroCrossingCount++;
                }
            }
            System.out.printf("%nFor M(x) with x from 1 to %,d%n", nMax);        
            System.out.printf("The maximum of M(x) is M(%,d) = %,d.%n", mMaxIndex, mMax);
            System.out.printf("The minimum of M(x) is M(%,d) = %,d.%n", mMinIndex, mMin);
            System.out.printf("The sum of M(x) is %,d.%n", mSum);
            System.out.printf("The count of positive M(x) is %,d, count of negative M(x) is %,d.%n", positiveCount, negativeCount);
            System.out.printf("M(x) has %,d zeroes in the interval.%n", zeroCount);
            System.out.printf("M(x) has %,d crossings in the interval.%n", zeroCrossingCount);
        }
    }
    
    private static int MU_MAX = 100_000_000;
    private static int[] MU = null;
    private static int[] MERTEN = null;
        
    
    private static int mertenFunction(int n) {
        if ( MERTEN != null ) {
            return MERTEN[n];
        }
        
        
        MU = new int[MU_MAX+1];
        MERTEN = new int[MU_MAX+1];
        MERTEN[1] = 1;
        int sqrt = (int) Math.sqrt(MU_MAX);
        for ( int i = 0 ; i < MU_MAX ; i++ ) {
            MU[i] = 1;
        }
        
        for ( int i = 2 ; i <= sqrt ; i++ ) {
            if ( MU[i] == 1 ) {
                
                for ( int j = i ; j <= MU_MAX ; j += i ) {
                    MU[j] *= -i;
                }
                
                for ( int j = i*i ; j <= MU_MAX ; j += i*i ) {
                    MU[j] = 0;
                }
            }
        }
        
        int sum = 1;
        for ( int i = 2 ; i <= MU_MAX ; i++ ) {
            if ( MU[i] == i ) {
                MU[i] = 1;
            }
            else if ( MU[i] == -i ) {
                MU[i] = -1;
            }
            else if ( MU[i] < 0 ) {
                MU[i] = 1;               
            }
            else if ( MU[i] > 0 ) {
                MU[i] = -1;
            }
            sum += MU[i];
            MERTEN[i] = sum;
        }
        return MERTEN[n];
    }

}