%I #2 Mar 30 2012 17:23:31
%S 1,2,4,10,13,3,19,38,16,5,9,73,48,43,23,6,15,42,7,14,45,8,49,64,12,72,
%T 17,50,97,154,20,95,27,98,18,83,21,99,91,173,22,107,89,103,190,169,28,
%U 117,104,127,155,24,118,219,26,135,29,142,258,25,147,36,181,11,35,159
%N Lexicographically earliest permutation of the positive integers such that the inverse permutation is also the absolute value of the first differences.
%e Let a(n) be this sequence and b(n)=|a(n)-a(n+1)| be the inverse permutation of this sequence.
%e After a(1)=1, a(2)=2, a(3)=4, the next term, a(4), cannot be a repeat of 1,2, or 4 since by definition a(n) must be a permutation of the positive integers.
%e It cannot be 3,5, or 6, as that would force b(3)=1 or 2 (a repeat of b(1)=1, or b(2)=2).
%e We cannot have a(4)=7, because b(3)=3 implies a(3)=3, which contradicts a(3)=4.
%e We cannot have a(4)=8, because b(3)=4 implies a(4)=3.
%e We cannot have a(4)=9, because b(3)=5 implies a(5)=3, and b(4)=|a(5)-a(4)|=6 which contradicts b(4)=3 as implied by a(3)=4.
%e Therefore a(4)=10 is the smallest value of a(4) which will not generate a contradiction.
%Y Cf. A180428 - Inverse Permutation of this sequence. Also the first differences (absolute value) of this sequence.
%K nice,nonn
%O 1,2
%A _Andrew Weimholt_, Sep 04 2010