List files

Check out this link

If you are generating the files and you can choose a file name my suggestion is to pad the number with zeros. The following function helps with that:

numstring(x, n=3) = string(x + 10^n)[2:end]

So in your case,

julia> fnames = "file_" .*  numstring.(1:13, 3)
13-element Array{String,1}:
 "file_001"
 "file_002"
 "file_003"
 "file_004"
 "file_005"
 "file_006"
 "file_007"
 "file_008"
 "file_009"
 "file_010"
 "file_011"
 "file_012"
 "file_013"

In the link above, Palli suggests the following procedure:

julia> split(read(`ls -1v`, String))
19-element Array{SubString{String},1}:
 "file_1" 
 "file_2" 
 "file_3" 
 "file_4" 
 "file_5" 
 "file_6" 
 "file_7" 
 "file_8" 
 "file_9" 
 "file_10"
 "file_11"
 "file_12"
 "file_13"
 "file_14"
 "file_15"
 "file_16"
 "file_18"
 "file_20"
 "file_25"
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