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%I #19 Nov 26 2014 21:10:51
%S 0,0,0,2,4,6,9,11,16,19,22,26,32,38,43,50,56,67,75,81,89,97,109,119,
%T 130,140,154,166,178,194,205,220,233,250,264,283,296,312,327,345,359,
%U 378,397,415,432,456,481,504,523,547,569,591,617,641,664,689,718,744,769,797,824,847,878,910,945
%N a(0) = 0. a(n) is the number of repeating sums in the collection of all sums of two elements in [a(0), ... a(n-1)], chosen without replacement.
%C Without replacement means that a(i)+a(i) is not a valid sum to include. However, if a(i) = a(j), a(i)+a(j) is still a valid sum to include because they have different indices.
%C a(i)+a(j) and a(j)+a(i) are regarded as the same sum for all indices i and j.
%C a(n) <= A000217(n)-n.
%e a(1) gives the number of repeating sums in the collection of all possible sums of two elements in [0]. There are no sums between two elements here, so a(1) = 0.
%e a(2) gives the number of repeating sums in the collection of all possible sums of two elements in [0,0]. There is only one sum, 0, thus there are no repeats. So a(2) = 0.
%e a(3) gives the number of repeating sums in the collection of all possible sums of two elements in [0,0,0]. The possible sums are 0+0, 0+0, or 0+0, thus there are two repeats. So a(3) = 2.
%e a(4) gives the number of repeating sums in the collection of all possible sums of two elements in [0,0,0,2]. The possible sums are 0+0, 0+0, 0+2, 0+0, 0+2, and 0+2. There are 4 repeating sums (2 extra zeros and 2 extra twos). So a(4) = 4.
%o (PARI) v=[0];n=1;while(n<75,w=[];for(i=1,#v,for(j=i+1,#v,w=concat(w,v[i]+v[j])));v=concat(v,#w-#vecsort(w,,8));n++);v
%Y Cf. A247184.
%K nonn
%O 0,4
%A _Derek Orr_, Nov 22 2014
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