Re: Partition prime list into equal k sublists. How to

• To: mathgroup at smc.vnet.net
• Subject: [mg107047] Re: [mg107000] Partition prime list into equal k sublists. How to
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Mon, 1 Feb 2010 06:13:51 -0500 (EST)
• References: <201001311057.FAA10768@smc.vnet.net>

```Umm... that's the SUM of a list.

Bobby

On Sun, 31 Jan 2010 19:34:01 -0600, a boy <a.dozy.boy at gmail.com> wrote:

> I mean Total[list] as the "summary" of a list
>
> On Mon, Feb 1, 2010 at 5:09 AM, DrMajorBob <btreat1 at austin.rr.com> wrote:
>
>> What's the "summary" of a list?
>>
>> Bobby
>>
>>
>> On Sun, 31 Jan 2010 04:57:05 -0600, a boy <a.dozy.boy at gmail.com> wrote:
>>
>>  Suppose p[i] is the i-th prime,  P[n]={p[i]| 1<=i<=n}. Since the only
>> even
>>> prime is 2, the summary of P[n] is even iff n is odd.
>>>
>>> Conjecture: When n=2m+1 is odd, prime list P[n] can be partitioned
>>> into 2
>>> non-overlapping sublists , each sublist has equal summary
>>> Total[P[n]]/2;
>>> When n=2m is even, prime list P[n] can be partitioned into 2
>>> non-overlapping
>>> sublists , one sublist's summary  is (Total[P[n]]-1)/2, the other's is
>>> (Total[P[n]]+1)/2.
>>>
>>> k = 2;
>>> Manipulate[P[n] = list = Prime[Range[1, n]];
>>>  Print[sum = Total[list]/k];
>>>  Select[Subsets[list, {(n - 1)/2}], Total[#] == sum &],
>>>  {n, 3, 21, 2}]
>>>
>>>  n=10, P[10]=Prime[Range[1, 10]] can be partitioned into equal 3
>>> sublists.
>>> 43=129/3=3+11+29=7+13+23=2+ 5+17+ 19
>>> Question: when prime list can be partitioned into equal 3 sublists?
>>> only
>>> if
>>> Total[P[n]]/3 is an integer?
>>>
>>> n = 20
>>> FoldList[Plus, 0, Prime[Range[1, n]]]
>>> k = 3;
>>> n = 10;
>>> P[n] = list = Prime[Range[1, n]]
>>> sum = Total[list]/k
>>> Select[Subsets[list, {2, (n - 1)}], Total[#] == sum &]
>>>
>>> These codes is not good for solving this question. Can you help me?
>>>
>>>
>>>
>>
>> --
>> DrMajorBob at yahoo.com
>>

--
DrMajorBob at yahoo.com

```

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