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Actually, the loop only needs to go to floor(sqrt(n)), because once you have found one factor, it is obviously trivial to calculate another. This makes for a big improvement in speed for large numbers.
Many thanks for your time and help to my original post, Valery and Richard. 
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Best Regards.
Loi.
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Is there an other way ? (for 2015201620162015)
Yes. One could be to use symbolic eval instead of numeric, but I guess this will take a lot of time.
The second is this one as the mod-function seems to be a little more stable in this case. But I guess that sooner or later you will run into troubles with this method, too. After all the magnitude of numbers IS limited if you use the numerics.
Some remarks:
1) You will get a wrong result if x is a square of an integer as this routine then will count that integer twice. Should e easy to fix.
2) The usual convention in number theory is to add 1 and x to the list of divisors of a number. Even easier to fix.
3) As you just want to see the total number of divisors its not very efficient to create the whole vector (and then even sort it), just use a variable which counts. Should also be easy to implement.
4) As you probably know there is a much more efficient algorithm to get the number you want if we know the prime factors of x.
If the primefactorisation of x is
then the searched for number of different factors of x is
Example: 2015201620162015 is the product of four primes, as you have shown. So N=4 and each n.k is 1. The total number of factors (including 1, and x) therefore is (1+1)^4=16.
Unfortunately we do not have access to the results of the symbolic "factor" and so you would have to write your own routines which does the prime factorisation and counts how often each prime factor is used. You will run into numerical inaccuracies here, too, if x is big enough, I guess, so probably its not worth a try.
WE
You will run into numerical inaccuracies here, too, if x is big enough, I guess, so probably its not worth a try.
It may not be worth the effort, but it sure is a nice pastime.
So here is a routine using the algorithm described above.
I am pretty sure you could speed up the local routine f() which does the primefactorisation recursively using the usual methods (don't try divison by numbers already tried, make it iterative, use arrays with precalculated primes, ...)
But you sure run into numerical problems with slightly higher numbers:
So I was correct - not worth the effort 😉
WE
The second is this one as the mod-function seems to be a little more stable in this case.
The limit of 12 digits is because "Use exact equality for comparisons and truncation" is unchecked in the worksheet options. Checking that would allow more digits, but at the risk of other numerical problems elsewhere in the worksheet. (From the help: "When not checked, the absolute value of the difference between two numbers divided by their average must be <10-12 for them to be considered equal, and only the first 12 decimal places are used in truncation."). Using the mod function is a good way to get around this.
1) You will get a wrong result if x is a square of an integer as this routine then will count that integer twice. Should e easy to fix.
Good catch. This fixes it:
2) The usual convention in number theory is to add 1 and x to the list of divisors of a number. Even easier to fix.
Well, I did say it calculated the proper factors. As you say, it is easy enough to add the trivial ones, but I'm not going to do it 
Here's a shorter and iterative routine counting the number of factors.
Of course, as the last examples show, we run into the same numerical limitations.
Symbolic eval seems to run endless - I canceled calculation.
Relative query:
Is it possible to create a program function for Unique Function with Input_Output as the above ? ( for Letter, Not number )
Best Regards.
1). No. The factor keyword returns the number as the product of prime factors. Those are not all the factors (or even necessarily all the prime ones).
2). If the letters are strings, not undefined variable names:
This assumes the vector is sorted.
Richard Jackson написал(а):
You only need the loop to go from 2 to floor(n/2). 1 and n are factors, but they are not proper factors (and can be easily added outside the loop if desired).
I thought those extra math symbols provided in Prime were just cosmetic. I didn't realize any of them actually "worked". I can't find anything about them in the help. Do you know if they all work, and if so what does a symbol such as "because" does?
KISS - keep it simple...
