

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




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

Table of n, a(n) for n=0..7.
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



