R Function for returning ALL factors

To follow up on my comment (thanks to @Ramnath for my typo), the brute force method seems to work reasonably well here on my 64 bit 8 gig machine:

FUN <- function(x) {
    x <- as.integer(x)
    div <- seq_len(abs(x))
    factors <- div[x %% div == 0L]
    factors <- list(neg = -factors, pos = factors)
    return(factors)
}

A few examples:

> FUN(100)
$neg
[1]   -1   -2   -4   -5  -10  -20  -25  -50 -100

$pos
[1]   1   2   4   5  10  20  25  50 100

> FUN(-42)
$neg
[1]  -1  -2  -3  -6  -7 -14 -21 -42

$pos
[1]  1  2  3  6  7 14 21 42

#and big number

> system.time(FUN(1e8))
   user  system elapsed 
   1.95    0.18    2.14 

You can get all factors from the prime factors. gmp calculates these very quickly.

library(gmp)
library(plyr)

get_all_factors <- function(n)
{
  prime_factor_tables <- lapply(
    setNames(n, n), 
    function(i)
    {
      if(i == 1) return(data.frame(x = 1L, freq = 1L))
      plyr::count(as.integer(gmp::factorize(i)))
    }
  )
  lapply(
    prime_factor_tables, 
    function(pft)
    {
      powers <- plyr::alply(pft, 1, function(row) row$x ^ seq.int(0L, row$freq))
      power_grid <- do.call(expand.grid, powers)
      sort(unique(apply(power_grid, 1, prod)))
    }
  )
}

get_all_factors(c(1, 7, 60, 663, 2520, 75600, 15876000, 174636000, 403409160000))