It seems that you mighg want to use a diferent name for your combE

load 'stats'
   
   
   
   (3 comb 4);3 combE 4
┌─────┬───────┐
│0 1 2│0 0 0 1│
│0 1 3│0 0 0 2│
│0 2 3│0 0 1 1│
│1 2 3│0 0 1 2│
│     │0 0 2 1│
│     │0 0 2 2│
│     │0 1 0 1│
│     │0 1 0 2│
│     │0 1 1 1│
│     │0 1 1 2│
│     │0 1 2 1│
│     │0 1 2 2│
│     │0 2 0 1│
│     │0 2 0 2│
│     │0 2 1 1│
│     │0 2 1 2│
│     │0 2 2 1│
│     │0 2 2 2│
│     │1 0 0 1│
│     │1 0 0 2│
│     │1 0 1 1│
│     │1 0 1 2│
│     │1 0 2 1│
│     │1 0 2 2│
│     │1 1 0 1│
│     │1 1 0 2│
│     │1 1 1 1│
│     │1 1 1 2│
│     │1 1 2 1│
│     │1 1 2 2│
│     │1 2 0 1│
│     │1 2 0 2│
│     │1 2 1 1│
│     │1 2 1 2│
│     │1 2 2 1│
│     │1 2 2 2│
└─────┴───────┘
   
Linda


-----Original Message-----
From: Programming [mailto:[email protected]] On Behalf 
Of Erling Hellenäs
Sent: Friday, November 3, 2017 12:44 PM
To: [email protected]
Subject: Re: [Jprogramming] Partitions

Hi all!

I did not correctly understand the e. verb

    24 e. r4
0
    24 e. ,r4
1

So 358358 surely was there and was double-counted.

Cheers,

Erling Hellenäs

On 2017-11-03 17:07, Erling Hellenäs wrote:
> No, its not. It seems it should have bin there. Maybe something is 
> wrong. /Erling
>
> On 2017-11-03 16:56, Erling Hellenäs wrote:
>> Lol. This root is double-counted. 1 1 1 1 358358 /Erling
>>
>> On 2017-11-03 16:42, Erling Hellenäs wrote:
>>> Hi all!
>>>
>>> My take:
>>>
>>>    v=:1 1 1 1 2 7 11 13 179
>>>    r=:5 parRuskeyE 9
>>>    r2=: >r (*/)@:{&.>"1 0 < v
>>>    r3=:/:~"1 r2
>>>    r4=: ~.r3
>>>    NB. One root is 1 1 1 1 358358
>>>    NB. All permutations
>>>    (!5)*1+#r4
>>> 6360
>>>
>>> Cheers,
>>>
>>> Erling Hellenäs
>>>
>>> On 2017-11-03 15:03, Erling Hellenäs wrote:
>>>> Sorry. Add ones. /Erling
>>>>
>>>> On 2017-11-03 15:00, Erling Hellenäs wrote:
>>>>> OK. The five prime factors are needed if you want to multiply 5 
>>>>> numbers and get 358358, except that you can possibly add zeros and  
>>>>> 358358 itself? /Erling
>>>>>
>>>>> On 2017-11-03 14:35, 'Mike Day' via Programming wrote:
>>>>>> Language differences?
>>>>>>
>>>>>> Personally,  I tend to be rather careless,  using the terms 
>>>>>> “factor” and “divisor” for the same thing,  when “divisor” is 
>>>>>> perhaps to be preferred.  However,  Pari GP provides a library 
>>>>>> function “factor” to deliver the prime factorisation of its 
>>>>>> argument,  in a similar fashion to J’s q: .
>>>>>>
>>>>>> So, strictly speaking,  perhaps,  the factors of 358358 are 2 7
>>>>>> 11 13 179 and its divisors are 1 2 7 11 13 14 22 etc...
>>>>>>
>>>>>> You’re both right!
>>>>>>
>>>>>> Mike
>>>>>>
>>>>>> Please reply to [email protected].
>>>>>> Sent from my iPad
>>>>>>
>>>>>>> On 3 Nov 2017, at 11:25, Erling Hellenäs 
>>>>>>> <[email protected]> wrote:
>>>>>>>
>>>>>>> Hi all!
>>>>>>>
>>>>>>> Raul:
>>>>>>>
>>>>>>> "Hmm... actually, thinking about it, the par approach here is 
>>>>>>> not efficient enough for this example. 5 parRuskeyE 32 is too 
>>>>>>> big of a result, I think. (358358 has 5 distinct prime factors 
>>>>>>> and, thus, 32 integer factors.)"
>>>>>>>
>>>>>>> However there are only 5 integer factors:
>>>>>>>
>>>>>>>    q: 358358
>>>>>>> 2 7 11 13 179
>>>>>>>    */q:358358
>>>>>>> 358358
>>>>>>>
>>>>>>> Now you ask me for 44 5 integer factorizations.
>>>>>>>
>>>>>>> As far as I understand there is only one 5 integer factorization 
>>>>>>> of 358358 unless you count 1 and 358358 as factors.
>>>>>>> In any case it can be handled by parRuskeyE.
>>>>>>>
>>>>>>> Cheers,
>>>>>>>
>>>>>>> Erling Hellenäs
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>> Den 2017-11-03 kl. 10:58, skrev Raul Miller:
>>>>>>>> I'm not sure where you showed the 44 different 5 integer 
>>>>>>>> factorizations of 358358?
>>>>>>>>
>>>>>>>> Thanks,
>>>>>>>>
>>>>>>> ----------------------------------------------------------------
>>>>>>> ------
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>>>>>
>>>>>
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>>>>
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