This interpretation is correct; '5=2+3 19=17+2, 11+5+3 31=29+2, 23+5+3, 19+7+3+2 43=41+2, 31+7+5, 29+7+5+2, 17+11+7+5+3 61=59+2, 31+19+11, 31+17+11+2, 23+19+11+5+3, 23+17+11+5+3+2 73=71+2, 61+7+5, 59+7+5+2, 31+19+11+7+5, 31+19+11+7+3+2' although 73 as listed has only 6 distinct primes, and to qualify requires 7; Two other interesting sets of integers are the Fibonacci numbers and the Partition numbers. With Fibonacii, let a_1=1 and a_2=1, and these are distinct, so: 2=1+1 5=3+2=3+1+1 13=8+5=8+3+2=8+3+1+1 34=21+13=21+8+5=21+8+3+2=21+8+3+1+1 so there is an obvious pattern here. The Partition numbers seem to behave as if they avoid being the sum of 2 distinct previous elements (we can even drop the distinct part) after 30. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com -----Original Message----- From: math-fun-bounces+perry=globalnet.co.uk@mailman.xmission.com [mailto:math-fun-bounces+perry=globalnet.co.uk@mailman.xmission.com]On Behalf Of Gershon Bialer Sent: 12 January 2004 00:30 To: math-fun Subject: Re: [math-fun] Subset Problem I think he wants the number to be the sum of 1 distinct element, 2 distinc elements, 3 distinct elements,,..., and k distinct elements. Just because a,b \in S does not imply a+b \in S. So, just because k+1 distinct elements of S add up to something doesn't not mean k elements of S add up to something. For example there is no number which is the of both 2 and 1 distinct elements of {1,2,4,8,16} or the subset of any other geometric progression for that matter. You can just get the sets of elements that are the sum 1 distinct elments, 2 distinct elements, ..., and take their interesection. If all of the integers in S are positive, then you can order them, and see if S_m+S_m+1+..S_m+k<S_(m+k+1) for all positive integers m with a fixed k in which case you could easily prove that there are no numbers such that 1,2,3,...,k distinct elements of S sum up to a given number. It might be interesting let S be the set of primes, and consider Perry's question. 5=2+3 19=17+2, 11+5+3 31=29+2, 23+5+3, 19+7+3+2 43=41+2, 31+7+5, 29+7+5+2, 17+11+7+5+3 61=59+2, 31+19+11, 31+17+11+2, 23+19+11+5+3, 23+17+11+5+3+2 73=71+2, 61+7+5, 59+7+5+2, 31+19+11+7+5, 31+19+11+7+3+2 Note: I just found these by hand If p is (k-1)'th number satisfying i) p is prime ii) p does not equal 7 iii) p=q+2 where q is prime then is p always equal to the sum of n distinct primes where n<=k? Are the representations as sums of k primes unique? I think generalizations of these thing could relate to quesitons like how many solutions are there to the equations p+q=r+s for p,q,r,s distinct primes? Gershon Bialer ----- Original Message ----- From: "Allan C. Wechsler" <acw@alum.mit.edu> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Sunday, January 11, 2004 4:55 PM Subject: Re: [math-fun] Subset Problem
At 11:09 PM 1/8/04 +0000, you wrote:
If we let S be a set of n integers, then how can we determine the lowest number x_k such that x_k is a sum of 1, 2, 3, ..., k distinct elements of S?
I must not be understanding the problem. What is wrong with simply adding up the k smallest elements of S? -A
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