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

• To: mathgroup at smc.vnet.net
• Subject: [mg107033] Re: [mg107000] Partition prime list into equal k sublists. How to
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Mon, 1 Feb 2010 06:11:17 -0500 (EST)
• References: <201001311057.FAA10768@smc.vnet.net>

```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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