%I
%S 0,1,1,1,1,1,2,1,2,1,3,1,3,2,1,5,2,4,4,2,1,6,3,5,6,4,2,1,10,5,7,9,7,4,
%T 2,1,13,7,9,11,11,7,4,2,1,19,11,12,15,16,12,7,4,2,1,25,16,15,19,22,18,
%U 12,7,4,2,1,35,24,20,26,29,27,19,12,7,4,2,1
%N Triangle read by rows: T(n,k) = number of partitions of n with positive kth rank.
%C We define the kth rank of a partition as the kth part minus the number of parts >= k. Every partition of n has n ranks. This is a generalization of the Dyson's rank of a partition which is the largest part minus the number of parts. Since the first part of a partition is also the largest part of the same partition so the Dyson's rank of a partition is the case for k = 1.
%C It appears that the sum of the kth ranks of all partitions of n is equal to zero.
%C Also T(n,k) = number of partitions of n with negative kth rank.
%C It appears that reversed rows converge to A000070, the same as A208482.  Omar E. Pol, Mar 11 2012
%H Alois P. Heinz, <a href="/A208478/b208478.txt">Rows n = 1..44, flattened</a>
%e For n = 4 the partitions of 4 and the four types of ranks of the partitions of 4 are
%e 
%e Partitions First Second Third Fourth
%e of 4 rank rank rank rank
%e 
%e 4 41 = 3 01 = 1 01 = 1 01 = 1
%e 3+1 32 = 1 11 = 0 01 = 1 00 = 0
%e 2+2 22 = 0 22 = 0 00 = 0 00 = 0
%e 2+1+1 23 = 1 11 = 0 10 = 1 00 = 0
%e 1+1+1+1 14 = 3 10 = 1 10 = 1 10 = 1
%e 
%e The number of partitions of 4 with positive kth ranks are 2, 1, 2, 1 so row 4 lists 2, 1, 2, 1.
%e Triangle begins:
%e 0;
%e 1, 1;
%e 1, 1, 1;
%e 2, 1, 2, 1;
%e 3, 1, 3, 2, 1;
%e 5, 2, 4, 4, 2, 1;
%e 6, 3, 5, 6, 4, 2, 1;
%e 10, 5, 7, 9, 7, 4, 2, 1;
%e 13, 7, 9, 11, 11, 7, 4, 2, 1;
%e 19, 11, 12, 15, 16, 12, 7, 4, 2, 1;
%e 25, 16, 15, 19, 22, 18, 12, 7, 4, 2, 1;
%e 35, 24, 20, 26, 29, 27, 19, 12, 7, 4, 2, 1;
%Y Column 1 is A064173. Row sums give A208479.
%Y Cf. A063995, A105805, A181187, A194547, A194549, A195822, A208482, A209616.
%K nonn,tabl
%O 1,7
%A _Omar E. Pol_, Mar 07 2012
%E More terms from _Alois P. Heinz_, Mar 11 2012
