Question

I want to make the code below fast. It takes so long time to run, and I got this error:

Warning: FOR loop index is too large. Truncating to 2147483647.

I need to calculate over 3^100 so… is it impossible?

function sodiv = divisorSum(n)
    sodiv = 0;
    for i=1:n
        if (mod(n,i) == 0)
            sodiv = sodiv + i;
        end
    end
end

function finalSum1 = formular1(N,n)
    finalSum1 = 0;
    for k = 1:N
       finalSum1 = finalSum1 + (divisorSum(k) * divisorSum(3^n*(N-k)));
    end
end

Nv=100;
nv=[1:20];

for i=1:length(nv)
    tic;
    nfunc1(i)=formular1(Nv,nv(i));
    nt1(i)=toc;
    sprintf('nt1 : %d finished, %f', i,nt1(i))
end

The purpose of this code is to check the algorithm’s calculation time.

Answer 1

This code will never finish, because it is so inefficient.

For instance, there is a function that counts number of all divisors and is going through all numbers from 1 to N and count. But using an efficient formula would make it run much master.

Let’s say that one need to sum divisors of number a^b where a is prime number.
Instead of calculating a^b and going form 1 to a^b, one can see that it is better going
a^1, a^2, a^3, ..., a^n, because only these numbers are divisors. But you can go even further and observe that the sum of these numbers are the sum of geometric progression so the number of divisors become:

sum divisors, a^b = (a^(b+1)-1) / (a-1)