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Alternating sum of compositions in standard order.
71

%I #5 Jul 24 2013 18:25:30

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

%T 1,6,4,2,4,0,2,4,2,-2,0,2,0,4,2,0,2,-4,-2,0,-2,2,0,-2,0,4,2,0,2,-2,0,

%U 2,0,7,5,3,5,1,3,5,3,-1,1,3,1,5,3,1,3,-3,-1,1,-1,3,1,-1,1,5,3,1,3,-1,1,3,1,-5,-3,-1,-3,1,-1,-3,-1,3

%N Alternating sum of compositions in standard order.

%C The standard order of compositions is given by A066099.

%C The sum of row n is 2^{n-1} for n>0.

%H Alois P. Heinz, <a href="/A124754/b124754.txt">Rows n = 0..14, flattened</a>

%F For a composition b(1),...,b(k), a(n) = Sum_{i=1}^k (-1)^{i-1} b(i).

%F a(2^k) = k+1. If n = 2^e_1 + 2^e_2 + k, 0 <= k < 2^e_2 < 2^e_1, then a(n) = (e_1 - e_2) - a(2^e_2 + k).

%F a(0) = 0; for n>0, a(n) = a(floor(n/2)) - A106400(n).

%e Composition number 11 is 2,1,1; 2-1+1 = 2, so a(11) = 2.

%e The table starts:

%e 0

%e 1

%e 2 0

%e 3 1 -1 1

%Y Cf. A066099, A070939, A124756, A011782 (row lengths), A106400.

%K easy,sign,tabf

%O 0,3

%A _Franklin T. Adams-Watters_, Nov 06 2006