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%I #19 Dec 31 2020 11:11:15
%S 1,5,14,1,30,2,55,6,91,10,1,140,19,2,204,28,3,285,44,7,385,60,11,1,
%T 506,85,15,2,650,110,24,3,819,146,33,4,1015,182,42,8,1240,231,58,12,1,
%U 1496,280,74,16,2,1785,344,90,20,3,2109,408,115,29,4,2470,489,140,38,5,2870,570,165,47,9,3311,670,201,56,13,1
%N Triangle read by rows T(n,k) in which column k lists the partial sums of the k-th column of triangle A236104.
%C Alternating sum of row n equals A175254(n), i.e., sum_{k=1..A003056(n))} (-1)^(k-1)*T(n,k) = A175254(n), which is also the volume (or the total number of units cubes) in the first n levels of the stepped pyramid described in A245092.
%C Row n has length A003056(n) hence the first element of column k is in row A000217(k).
%e Triangle begins:
%e 1;
%e 5;
%e 14, 1;
%e 30, 2;
%e 55, 6;
%e 91, 10, 1;
%e 140, 19, 2;
%e 204, 28, 3;
%e 285, 44, 7;
%e 385, 60, 11, 1;
%e 506, 85, 15, 2;
%e 650, 110, 24, 3;
%e 819, 146, 33, 4;
%e 1015, 182, 42, 8;
%e 1240, 231, 58, 12, 1;
%e 1496, 280, 74, 16, 2;
%e 1785, 344, 90, 20, 3;
%e 2109, 408, 115, 29, 4;
%e 2470, 489, 140, 38, 5;
%e 2870, 570, 165, 47, 9;
%e 3311, 670, 201, 56, 13, 1;
%e 3795, 770, 237, 72, 17, 2;
%e 4324, 891, 273, 88, 21, 3;
%e 4900, 1012, 322, 104, 25, 4;
%e ...
%e For n = 6 we have that A175254(6) = [1] + [1 + 3] + [1 + 3 + 4] + [1 + 3 + 4 + 7] + [1 + 3 + 4 + 7 + 6] + [1 + 3 + 4 + 7 + 6 + 12] = 1 + 4 + 8 + 15 + 21 + 33 = 82. On the other hand the alternating sum of the 6th row of the triangle is 91 - 10 + 1 = 82, equaling A175254(6).
%Y Cf. A000203, A000217, A003056, A024916, A175254, A196020, A235791, A236104, A237048, A237591, A237593, A237270, A237271, A245092, A261699, A262626.
%Y Column 1 gives A000330, n >= 1. Column 2 is A005993. It appears that column 3 is A092353.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Nov 03 2015
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