# Lower prime divisor

Is there a way in Sage to define the LOWER PRIME FACTOR of n without calculating ALL prime divisors with the list prime_factors(n) ? I would like to make that more quickly Thanks

Lower prime divisor

**
asked 2014-11-24 15:29:30 +0200 **

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Is there a way in Sage to define the LOWER PRIME FACTOR of n without calculating ALL prime divisors with the list prime_factors(n) ? I would like to make that more quickly Thanks

2

There a two ways to avoid calculating all prime divisors. Firstly, look at "trial_division". It gives the first prime divisor.

```
trial_division(13*97)
13
```

Secondly, you can use n.factor(limit=yoursearchlimit).

```
(13*97*191).factor(limit=50)
13 * 18527
```

The problem is that I look for the lower factor, which I don't know how large can be. So I cannnot impose a search limit. The only thing that I want is that when it find it, it stops searching for the other prime factors. Ok, that can be done by trial_division(n) (thank you for telling me that), but I suppose that it can be much slower than prime_factors(n) for large n. I hoped that there was something like "lower_primefactor(n)" already in sage... :)

Actually `trial_division`

is exactly what you are asking for, if you don't like the name you can do `lower_prime_factor = trial_division`

and then call `lower_prime_factor(n)`

. You can check the documentation and source code for `trial_division`

by doing `a = 3`

and then `a.trial_division?`

and `a.trial_division??`

.

1

It depends on the numbers you are factoring. Factorization algorithms for large numbers do not find factors in increasing order, thus you need to compute all factors in order to know which is smallest.

If your number is small, or has very small factors, this might be fasterâ€¯:

```
sage: n = 123456789
sage: bound = 100
sage: next(p for p in primes(bound) if n % p == 0)
3
```

play around with `bound`

to find the value that works for you. Be aware that if no prime up to `bound`

divides `n`

, the last instruction will raise an exception, in that case you can try a higher bound, or use `prime_divisors`

.

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Asked: ** 2014-11-24 15:29:30 +0200 **

Seen: **1,025 times**

Last updated: **Nov 24 '14**

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