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%I #20 Dec 01 2013 13:35:01
%S 1,1,2,1,1,3,1,1,1,4,4,1,1,1,1,1,5,5,1,1,1,1,1,1,1,6,6,6,6,1,1,1,1,1,
%T 1,1,1,1,1,1,7,7,7,7,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,8,8,8,8,8,8,8,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
%N Triangle read by rows: T(n,k) = sum of all parts in the k-th zone of the last section of the set of partitions of n.
%C Row n lists A000041(n-1) 1's together with A002865(n) n's.
%e Illustration of three arrangements of the last section of the set of partitions of 7 and the zone numbers:
%e --------------------------------------------------------
%e Zone \ a) b) c)
%e --------------------------------------------------------
%e 15 (7) (7) (. . . . . . 7)
%e 14 (4+3) (4+3) (. . . 4 . . 3)
%e 13 (5+2) (5+2) (. . . . 5 . 2)
%e 12 (3+2+2) (3+2+2) (. . 3 . 2 . 2)
%e 11 (1) (1) (1)
%e 10 (1) (1) (1)
%e 9 (1) (1) (1)
%e 8 (1) (1) (1)
%e 7 (1) (1) (1)
%e 6 (1) (1) (1)
%e 5 (1) (1) (1)
%e 4 (1) (1) (1)
%e 3 (1) (1) (1)
%e 2 (1) (1) (1)
%e 1 (1) (1) (1)
%e .
%e For n = 7 and k = 12 we can see that in the 12th zone of the last section of 7 the parts are 3, 2, 2, therefore T(7,12) = 3+2+2 = 7.
%e Written as a triangle begins:
%e 1;
%e 1,2;
%e 1,1,3;
%e 1,1,1,4,4;
%e 1,1,1,1,1,5,5;
%e 1,1,1,1,1,1,1,6,6,6,6;
%e 1,1,1,1,1,1,1,1,1,1,1,7,7,7,7;
%e 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,8,8,8,8,8,8,8;
%Y Row n has length A000041(n). Row sums give A138879.
%Y Cf. A000041, A002865, A135010, A138121, A182284, A193173.
%K nonn,tabf
%O 1,3
%A _Omar E. Pol_, Apr 23 2012
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