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 A231085 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. 2
 1, 1, 1, 1, 2, 12, 168, 4676 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS 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. LINKS Vladimir Letsko, Mathematical Marathon, Problem 183 (in Russian) EXAMPLE Let a < b < c < d. There are two possible ways to arrange the sums in increasing order: 1) a+b < a+c < a+d < b+c < b+d < c+d, (for instance, a = 1, b = 3, c = 4, d = 5); 2) a+b < a+c < b+c < a+d < b+d < c+d, (for instance, a = 1, b = 2, c = 3, d = 5). Hence a(4) = 2. CROSSREFS Cf. A231074, A003121, A237749 Sequence in context: A052728 A052729 A098431 * A059522 A271857 A113149 Adjacent sequences:  A231082 A231083 A231084 * A231086 A231087 A231088 KEYWORD nonn,more AUTHOR Vladimir Letsko, Nov 03 2013 EXTENSIONS a(7) from Anton Nikonov, Feb 07 2014 Edited and a(0)=1 prepended by Max Alekseyev, Feb 19 2014 STATUS approved

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Last modified January 21 08:23 EST 2022. Contains 350475 sequences. (Running on oeis4.)