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%I A286312 #12 May 16 2017 00:13:45
%S A286312 1,2,1,3,3,1,4,5,3,1,5,7,6,5,1,6,9,9,6,6,1,7,11,12,10,7,7,1,8,13,15,
%T A286312 14,10,7,8,1,9,15,18,18,15,11,7,9,1,10,17,21,22,20,15,12,17,10,1
%N A286312 Table read by antidiagonals upwards: a(n, k) is the minimum c such that n sets with k elements each can be constructed from numbers 1 to c (inclusive) such that any two sets have exactly 1 common element.
%H A286312 Simon Bohnen, Extended table of values
%H A286312 Simon Bohnen, A Java program for generating the sequence
%H A286312 Simon Bohnen and other users, Further discussion and proof for the formula
%F A286312 n < k+2: a(n,k) = kn-(n(n-1))/2.
%e A286312 Top-left corner of the array:
%e A286312 1 1 1 1 1 1 ...
%e A286312 2 3 3 5 6 7 ...
%e A286312 3 5 6 6 7 7 ...
%e A286312 4 7 9 10 10 11 ...
%e A286312 5 9 12 14 15 15 ...
%e A286312 6 11 15 18 20 21 ...
%e A286312 : : : : : : '.
%e A286312 For n=3 and k=3 the best possible solution is 6, the three sets are:
%e A286312 S1 = {1, 2, 3}
%e A286312 S2 = {1, 4, 5}
%e A286312 S3 = {2, 4, 6}
%K A286312 nonn,tabl
%O A286312 1,2
%A A286312 _Simon Bohnen_, May 06 2017
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