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September 24th, 2007, 07:42 AM  #1 
Newbie Joined: Sep 2007 Posts: 2 Thanks: 0  hard question with subsets
We have n>=1 and for the every proper subset A of the set {1,2,...,n} the number w(A) is given as follows: If a1>a2>a3>...>ak are the all elements of set A, then w(A)=a1a2+a3...+(1)^(k+1)*ak. Calculate the sum of all 2^(n1) integers w(A). 
September 24th, 2007, 12:25 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
W({1}) = 1 W({1, 2}) = 1 W({2}) = 2 W({1, 2, 3}) = 2 W({2, 3}) = 1 W({1, 3}) = 2 W({3}) = 3 W({1, 2, 3, 4}) = 2 W({2, 3, 4}) = 3 W({1, 3, 4}) = 2 W({1, 2, 4}) = 3 W({3, 4}) = 1 W({2, 4}) = 2 W({1, 4}) = 3 W({4}) = 4 Just thinking, don't have anything yet. 
September 24th, 2007, 02:28 PM  #3 
Newbie Joined: Sep 2007 Posts: 2 Thanks: 0 
It seems for me that general solution will be S(w(A))=n2^(n1). For n=1 we have one subset S(w(A))=1*2^0=1. For n=2 three subsets and S(w(a))=4=2*2^1. For n=3 seven subsets and S(w(a))=1+1+2+2+1+2+3=12=3*2^(31). For n=4 fifteen subsets and S(w(a))=4+12+9+6+1=32=4*2^(41). And for n=5 this sum should be equal 5*2^(51)=80. Now I am trying to prove the above conjecture... 

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