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%I #34 Apr 03 2024 03:08:15
%S 1,1,1,1,2,12,168,4680
%N The number of possible ways to arrange the sums x_i + x_j (1 <= i < j <= n) of the items x_1 < x_2 <...< x_n in increasing order provided that all sums are different.
%C For n<=5, a(n) = A231074(n), but for n>5, a(n) < A231074(n). For instance, let n = 6 and a < b < c < d < e < f. Then the arrangement a+b <= a+c <= a+d <= a+e <= b+c <= b+d <= a+f <= b+e <= b+f <= c+d <= c+e <= c+f <= d+e <= d+f <= e+f is possible (e.g., for a = 1, b = 5, c = 9, d = 12, e=13, f = 16), while the same arrangement with "<" instead of "<=" is not possible.
%H Arseniy Akopyan et al., <a href="https://mathoverflow.net/q/186515">Number of orders of k-sums of n numbers</a>, MathOverflow, 2014.
%H Vladimir Letsko, <a href="http://www-old.fizmat.vspu.ru/doku.php?id=marathon:problem_183">Mathematical Marathon, Problem 183</a> (in Russian)
%e Let a < b < c < d. There are two possible ways to arrange the sums in increasing order:
%e 1) a+b < a+c < a+d < b+c < b+d < c+d, (for instance, a = 1, b = 3, c = 4, d = 5);
%e 2) a+b < a+c < b+c < a+d < b+d < c+d, (for instance, a = 1, b = 2, c = 3, d = 5).
%e Hence a(4) = 2.
%Y Cf. A231074, A003121, A237749
%K nonn,more
%O 0,5
%A _Vladimir Letsko_, Nov 03 2013
%E a(7) from _Anton Nikonov_, Feb 07 2014
%E Edited and a(0)=1 prepended by _Max Alekseyev_, Feb 19 2014
%E a(7) corrected by _Max Alekseyev_, Apr 02 2024
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