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%I #15 May 22 2012 12:02:18
%S 1,3,4,5,8,9,11,16,19,20,15,26,31,34,35,31,46,57,62,65,66,39,70,85,96,
%T 101,104,105,71,110,141,156,167,172,175,176,94,165,204,235,250,261,
%U 266,269,270,150,244,315,354,385,400,411,416,419,420,196,346
%N Triangle read by rows: T(n,k) = sum of all parts of the last k shells of n.
%C The set of partitions of n contains n shells (see A135010). It appears that the last k shells of n contain p(n-k) parts of size k, where p(n) = A000041(n). See also A182703.
%F T(n,k) = A066186(n) - A066186(n-k).
%F T(n,k) = Sum_{j=n-k+1..n} A138879(j).
%e For n = 5 the illustration shows five sets containing the last k shells of 5 and below we can see that the sum of all parts of in each set:
%e --------------------------------------------------------
%e . S{5} S{4-5} S{3-5} S{2-5} S{1-5}
%e --------------------------------------------------------
%e . The Last Last Last The
%e . last two three four five
%e . shell shells shells shells shells
%e . of 5 of 5 of 5 of 5 of 5
%e --------------------------------------------------------
%e .
%e . 5 5 5 5 5
%e . 3+2 3+2 3+2 3+2 3+2
%e . 1 4+1 4+1 4+1 4+1
%e . 1 2+2+1 2+2+1 2+2+1 2+2+1
%e . 1 1+1 3+1+1 3+1+1 3+1+1
%e . 1 1+1 1+1+1 2+1+1+1 2+1+1+1
%e . 1 1+1 1+1+1 1+1+1+1 1+1+1+1+1
%e . ---------- ---------- ---------- ---------- ----------
%e . 15 26 31 34 35
%e .
%e So row 5 lists 15, 26, 31, 34, 35.
%e .
%e Triangle begins:
%e 1;
%e 3, 4;
%e 5, 8, 9;
%e 11, 16, 19, 20;
%e 15, 26, 31, 34, 35;
%e 31, 46, 57, 62, 65, 66;
%e 39, 70, 85, 96, 101, 104, 105;
%e 71, 110, 141, 156, 167, 172, 175, 176;
%e 94, 165, 204, 235, 250, 261, 266, 269, 270;
%e 150, 244, 315, 354, 385, 400, 411, 416, 419, 420;
%Y Mirror of triangle A212001. Column 1 is A138879. Right border is A066186.
%Y Cf. A135010, A182703, A212000, A212010.
%K nonn,tabl
%O 1,2
%A _Omar E. Pol_, Apr 26 2012
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