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

• To: mathgroup at smc.vnet.net
• Subject: [mg107039] Re: [mg107000] Partition prime list into equal k sublists. How to
• From: a boy <a.dozy.boy at gmail.com>
• Date: Mon, 1 Feb 2010 06:12:23 -0500 (EST)
• References: <201001311057.FAA10768@smc.vnet.net>

```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=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
>

```

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