The solution is actually really easy when not dp. It is just the count of numbers that are divisible by (1ll<<i) and only by (1ll<<i) times 2(i-1). If you just draw 1-15 tree and try to solve for 1 15 you'll observe it. You just have to make these checks (i'm too lazy to explain it, but if you want just comment):

    for (int i = 30; i >= 1; i--) {
        int powerOf2 = 1ll << i;
        int ci = countMultiples(n, m, powerOf2);
        ci -= al;
        int r = m / powerOf2;
        int l = n / powerOf2;
        an += 2 * (i - 1) * ci;

        if (m % powerOf2 == 0 && (r % 2 != 0 || r == 0) && ci > 0) {// <-- These are the checks
            an -= (i - 1);
        }
        if (n % powerOf2 == 0 && (l % 2 != 0 || l == 0) && ci > 0) {
            an -= (i - 1);
        }
        al += ci;
    }