# Practical Number?

Bob Johnsonson

Greenhorn

Posts: 2

posted 7 years ago

The Wikipedia article is helpful, I think.

Use some integer factorization algorithm to factor the number you want, getting each prime factor and the exponent it has in the prime factorization.

Store the prime factors in an sorted list [p1,p2,p3,...,pj] and exponent [a1,a2,a3,...,aj].

Set an integer sigma to 1.

If the first prime factor is not 2, the number is not practical.

Then for each prime factor:

Multiply sigma by (p_i^a_i-1)/(p_i-1)

if sigma+1 is greater than the next largest prime factor, your number is not practical.

Else, increment i and continue again.

If that holds for every prime factor, then your number is indeed practical.

That method, assuming a decent factorization algorithm is used, would work better than the method described in the previous post, I think.

Use some integer factorization algorithm to factor the number you want, getting each prime factor and the exponent it has in the prime factorization.

Store the prime factors in an sorted list [p1,p2,p3,...,pj] and exponent [a1,a2,a3,...,aj].

Set an integer sigma to 1.

If the first prime factor is not 2, the number is not practical.

Then for each prime factor:

Multiply sigma by (p_i^a_i-1)/(p_i-1)

if sigma+1 is greater than the next largest prime factor, your number is not practical.

Else, increment i and continue again.

If that holds for every prime factor, then your number is indeed practical.

That method, assuming a decent factorization algorithm is used, would work better than the method described in the previous post, I think.

Campbell Ritchie

Marshal

Posts: 52600

119

Campbell Ritchie

Marshal

Posts: 52600

119

Bob Johnsonson

Greenhorn

Posts: 2

posted 7 years ago

I'm not sure I could explain

*why*that works, but it just takes what Wikipedia says about the condition of how the ith prime factor is less than or equal to the sum of the divisors of the first i-1 factors (raised to their exponent)+1. Since the divisor function is multiplicative, i.e. σ(a*b)=σ(a)σ(b) if a and b are relatively prime, one doesn't need to compute the whole thing each time, you can just multiply by σ(p_i-1^a_i-1) each iteration, saving a bit of calculation.
Campbell Ritchie

Marshal

Posts: 52600

119

posted 7 years ago

I am afraid James is likely to be too busy to help until about Wednesday or Thursday, but I passed a message to him yesterday.

I did say

I did say

. . . and my algorithm runs in exponential complexity O(2^. . .

So who said there was anything simple about this ?

You may find other algorithms simpler.

*n*) where*n*is the number of factors <*m*, so finding an algorithm which is less efficient than mine is quite an achievement I got it to work, however, in 20-something lines of code. When I tried 1536, however, which has 19 factors < 1536, it had >500000 combinations to search and it took about one-third second longer to print a result than smaller numbers.