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%I #24 Mar 13 2015 22:57:55
%S 1,1,2,1,2,3,1,3,3,5,1,3,4,5,7,1,4,5,7,7,11,1,4,6,8,9,11,15,1,5,7,11,
%T 10,15,15,22,1,5,9,12,13,17,19,22,30,1,6,10,16,15,22,21,29,30,42,1,6,
%U 12,18,19,25,26,32,38,42,56,1,7,14,23,22,33,29,41,42,54,56,77
%N Triangle read by rows: T(n,k) = total number of parts of all regions of the set of partitions of n whose largest part is k.
%C For the definition of "region" see A206437.
%C T(n,k) is also the number of parts that end in the k-th column of the diagram of regions of the set of partitions of n (see Example section).
%e For n = 5 and k = 3 the set of partitions of 5 contains two regions whose largest part is 3, they are third region which contains three parts [3, 1, 1] and the sixth region which contains only one part [3]. Therefore the total number of parts is 3 + 1 = 4, so T(5,3) = 4.
%e .
%e . Diagram Illustration of parts ending in column k:
%e . for n=5 k=1 k=2 k=3 k=4 k=5
%e . _ _ _ _ _ _ _ _ _ _
%e . |_ _ _ | _ _ _ |_ _ _ _ _|
%e . |_ _ _|_ | |_ _ _| _ _ _ _ |_ _|
%e . |_ _ | | _ _ |_ _ _ _| |_|
%e . |_ _|_ | | |_ _| _ _ _ |_ _| |_|
%e . |_ _ | | | _ _ |_ _ _| |_| |_|
%e . |_ | | | | _ |_ _| |_| |_| |_|
%e . |_|_|_|_|_| |_| |_| |_| |_| |_|
%e .
%e k = 1 2 3 4 5
%e .
%e The 5th row lists: 1 3 4 5 7
%e .
%e Triangle begins:
%e 1;
%e 1, 2;
%e 1, 2, 3;
%e 1, 3, 3, 5;
%e 1, 3, 4, 5, 7;
%e 1, 4, 5, 7, 7, 11;
%e 1, 4, 6, 8, 9, 11, 15;
%e 1, 5, 7, 11, 10, 15, 15, 22;
%e 1, 5, 9, 12, 13, 17, 19, 22, 30;
%e 1, 6, 10, 16, 15, 22, 21, 29, 30, 42;
%e 1, 6, 12, 18, 19, 25, 26, 32, 38, 42, 56;
%e 1, 7, 14, 23, 22, 33, 29, 41, 42, 54, 56, 77;
%Y Column 1 is A000012. Column 2 are the numbers => 2 of A008619. Row sums give A006128, n>=1. Right border gives A000041, n>=1. Second right border gives A000041, n>=1.
%Y Cf. A006128, A133041, A135010, A138137, A139582, A141285, A182377, A186114, A186412, A187219, A193870, A194446, A206437, A207779, A211978, A220517, A225598, A225600, A225610.
%K nonn,tabl
%O 1,3
%A _Omar E. Pol_, Aug 02 2013
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