Using the matlab to achieve the a(k)+a(j) when k is not equal to j in every stage

i want to write a code which is a(k)+a(j), but k can't be equal to j,then sum it every stage ,i mean

1. when k=1,sumj=a(k)+a(j)=a(1)+a(2)=3, sumj=a(k)+a(j)=a(1)+a(3)=4, sumj=a(k)+a(j)=a(1)+a(4)=5,b=3+4+5=12
2. when k=2 ,sumj=3,sumj=5,sumj=6,b=3+5+6=14
3. when k=3 ,sumj=4,sumj=5,sumj=7,b=4+5+7=16
4. when k=4 ,sumj=5,sumj=6,sumj=7,b=5+6+7=18

    a=[1 2 3 4]

    sumj=0;
    b=0;

    for k=1:4
        for j=1:4       
            if j~=k
               sumj = a(k)+ a(j)
            end    
        end
       b=b+sumj
    end

but this code will show

> sumj=3,sumj=5,sumj=6,b=5+6=11 (b shouldn't be 11,it should be 14)

> sumj=4,sumj=5,sumj=7,b=5+6+7=18 (b shouldn't be 18,it should be 16)

> sumj=5,sumj=6,sumj=7,b=5+6+7+7=25 (b shouldn't be 25,it should be 18)

How do i modify the code to let the result become what i want?

Answer 1

[Stephen23]

a = [1,2,3,4];
for k = 1:4
    b = 0;  
    for j = 1:4       
        if j~=k
            b = b + a(k) + a(j);
        end    
    end
    disp(b)
end

Displays:

b =  12
b =  14
b =  16
b =  18

Note that no loop is required:

>> [X,Y] = ndgrid(1:4);
>> Z = ~eye(4);
>> V = sum(a(X).*Z+a(Y).*Z,1)
V =
   12   14   16   18

[yang-En Hsiao]

About the first method you said,if i want to plus 1 to all b,that is ,the b answer will become

b=12+1=13
b=14+1=15
b=16+1=17
b=18+1=19

How do i rewrite the code? because i found that i can't just rewrite like this

c=disp(b)+1

the window will tell me that

Error using disp

Too many output arguments

[Stephen23]

disp(b+1)