Bicubic Interpolation Error in image

Question

I have implemented Bicubic Interpolation taking the reference of these slides. Slide number 118 gives the formulation to compute the Bicubic Interpolation.

I have implemented my code in the same way keeping P(dr-m) and P(n-dc) as:

P_dr = 1/6*( (max(d_r-m+2,0)^3) -4*(max(d_r-m+1,0)^3) -6*(max(d_r-m,0)^3) -4*(max(d_r-m-1,0)^3));
P_dc = 1/6*( (max(n-d_c+2,0)^3) -4*(max(n-d_c+1,0)^3) -6*(max(n-d_c,0)^3) -4*(max(n-d_c-1,0)^3));   

where the max function defines by Q(X) as defined in the slide.

But when I run the code, I get the following image:

.

But when I change my normalizing factor to 1/32, I get:

My original image is this:

Can someone point as to where I am going wrong?

The code snippet is as follows:

for i = 5:r_new-4
    for j = 5:c_new-4
        x = i/SF;
        y = j/SF;
        ro = round(x);
        co = round(y);
        d_r = x - ro; 
        d_c = y - co;
        temp = 0;
        for m = -1:2
            for n = -1:2
              P_r = 1/32*( (max(d_r-m+2,0)^3) - 4*(max(d_r-m+1,0)^3) - 6*(max(d_r-m,0)^3) - 4*(max(d_r-m-1,0)^3));
              P_c = 1/32*( (max(n-d_c+2,0)^3) - 4*(max(n-d_c+1,0)^3) - 6*(max(n-d_c,0)^3) - 4*(max(n-d_c-1,0)^3));
              temp = temp + I(ro+m,co+n)*P_r*P_c;
            end
        end
        I_new(i,j) = temp;
    end
end