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>
- Reply-to: drmajorbob at yahoo.com
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