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Column 3 of triangle A117162; equals the Moebius transform of A117163 (column 2 of A117162) preceded by a zero.
2

%I #6 Mar 30 2012 18:37:44

%S 0,0,1,-1,-2,-1,-1,2,-1,3,-1,2,2,1,-2,1,2,1,-1,0,1,1,-1,-2,2,-2,-1,2,

%T 3,1,-1,-3,-2,-2,3,-1,2,1,-5,0,2,-1,2,0,4,-1,-2,-1,-1,-4,-3,2,3,1,2,

%U -4,1,-2,0,0,-1,1,-1,1,1,2,2,0,0,-4,3,-1,-2,-2,-1,1,5,3,1,-4,-3,-3,-1,-2,-3,-2,-3,1,2,-4,1,2,3,4,2,3,4,1,4

%N Column 3 of triangle A117162; equals the Moebius transform of A117163 (column 2 of A117162) preceded by a zero.

%C The inverse Moebius transform of column 4 of A117162 equals

%C this sequence preceded by a zero.

%F Equals the Moebius transform of column 2 preceded by a zero:

%F [0,0,1,-1,-2,-1,-1,2,...] = Moebius([0, 0,1,-1,-2,0,-1,1,...]).

%Y Cf. A117162, A112682, A008683 (Moebius); A117163 (column 2).

%K sign

%O 1,5

%A _Wouter Meeussen_ and _Paul D. Hanna_, Mar 05 2006