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%I #29 Nov 30 2013 21:30:34
%S 2,0,2,1,0,1,0,1,0,1,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,1,
%T 0,1,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,0,1,0,0,
%U 0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,1,0,1
%N Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (2 + m).
%C Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 2. For further information see A182703 and A135010.
%F T(k,m) = A182703(2+m,k), with T(k,m) = 0 if k > 2+m.
%F T(k,m) = A194812(2+m,k).
%e Triangle begins:
%e 2,
%e 0, 2,
%e 1, 0, 1,
%e 0, 1, 0, 1,
%e 0, 0, 1, 0, 1,
%e 0, 0, 0, 1, 0, 1,
%e 0, 0, 0, 0, 1, 0, 1,
%e 0, 0, 0, 0, 0, 1, 0, 1,
%e 0, 0, 0, 0, 0, 0, 1, 0, 1,
%e 0, 0, 0, 0, 0, 0, 0, 1, 0, 1,
%e ...
%e For k = 1 and m = 1; T(1,1) = 2 because there are two parts of size 1 in the last section of the set of partitions of 3, since 2 + m = 3, so a(1) = 2. For k = 2 and m = 1; T(2,1) = 0 because there are no parts of size 2 in the last section of the set of partitions of 3, since 2 + m = 3, so a(2) = 0.
%Y Always the sum of row k = p(2) = A000041(n) = 2.
%Y The first (0-10) members of this family of triangles are A023531, A129186, this sequence, A194703-A194710.
%Y Cf. A135010, A138121, A182712-A182714, A194812.
%K nonn,tabl
%O 1,1
%A _Omar E. Pol_, Feb 05 2012