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%I #72 Mar 09 2022 01:15:36
%S 1,4,8,1,15,4,1,21,8,4,1,1,33,15,8,4,4,1,1,41,21,15,8,8,4,4,1,1,1,1,
%T 56,33,21,15,15,8,8,4,4,4,4,1,1,1,1,69,41,33,21,21,15,15,8,8,8,8,4,4,
%U 4,4,1,1,1,1,1,1,1,87,56,41,33,33,21,21,15,15,15,15,8,8,8,8
%N Irregular triangle read by rows T(n,k) in which row n has length A000041(n-1) and every column k is A024916, n >= 1, k >= 1.
%C T(n,k) is the number of cubic cells (or cubes) in the k-th level starting from the base of the tower described in A221529 whose largest side of the base is equal to n (see example). - _Omar E. Pol_, Jan 08 2022
%F T(n,k) = A024916(A336811(n,k)).
%F T(n,k) = Sum_{j=1..n} A339278(j,k). - _Omar E. Pol_, Jan 08 2022
%e Triangle begins:
%e 1;
%e 4;
%e 8, 1;
%e 15, 4, 1;
%e 21, 8, 4, 1, 1;
%e 33, 15, 8, 4, 4, 1, 1;
%e 41, 21, 15, 8, 8, 4, 4, 1, 1, 1, 1;
%e 56, 33, 21, 15, 15, 8, 8, 4, 4, 4, 4, 1, 1, 1, 1;
%e 69, 41, 33, 21, 21, 15, 15, 8, 8, 8, 8, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1;
%e ...
%e For n = 9 the length of row 9 is A000041(9-1) = 22.
%e From _Omar E. Pol_, Jan 08 2022: (Start)
%e For n = 9 the lateral view and top view of the tower described in A221529 look like as shown below:
%e _
%e 22 1 | |
%e 21 1 | |
%e 20 1 | |
%e 19 1 | |
%e 18 1 | |
%e 17 1 | |
%e 16 1 |_|_
%e 15 4 | |
%e 14 4 | |
%e 13 4 | |
%e 12 4 |_ _|_
%e 11 8 | | |
%e 10 8 | | |
%e 9 8 | | |
%e 8 8 |_ _|_|_
%e 7 15 | | |
%e 6 15 |_ _ _| |_
%e 5 21 | | |
%e 4 21 |_ _ _|_ _|_
%e 3 33 |_ _ _ _| | |_
%e 2 41 |_ _ _ _|_|_ _|_ _
%e 1 69 |_ _ _ _ _|_ _|_ _|
%e .
%e Level Row 9 Lateral view
%e k T(9,k) of the tower
%e .
%e _ _ _ _ _ _ _ _ _
%e |_| | | | | | | |
%e |_ _|_| | | | | |
%e |_ _| _|_| | | |
%e |_ _ _| _|_| |
%e |_ _ _| _| _ _|
%e |_ _ _ _| |
%e |_ _ _ _| _ _|
%e | |
%e |_ _ _ _ _|
%e .
%e Top view
%e of the tower
%e .
%e For n = 9 and k = 1 there are 69 cubic cells in the level 1 starting from the base of the tower, so T(9,1) = 69.
%e For n = 9 and k = 22 there is only one cubic cell in the level 22 (the top) of the tower, so T(9,22) = 1.
%e The volume of the tower (also the total number of cubic cells) represents the 9th term of the convolution of A000203 and A000041 hence it's equal to A066186(9) = 270, equaling the sum of the 9th row of triangle. (End)
%o (PARI) f(n) = numbpart(n-1);
%o T(n, k) = {if (k > f(n), error("invalid k")); if (k==1, return (n)); my(s=0); while (k <= f(n-1), s++; n--; ); 1+s; } \\ A336811
%o g(n) = sum(k=1, n, n\k*k); \\ A024916
%o row(n) = vector(f(n), k, g(T(n,k))); \\ _Michel Marcus_, Jan 22 2022
%Y Row sums give A066186.
%Y Row lengths give A000041.
%Y The length of the m-th block in row n is A187219(m), m >= 1.
%Y Cf. A350637 (analog for the stepped pyramid described in A245092).
%Y Cf. A000203, A024916, A196020, A221529, A236104, A235791, A237270, A237271, A237593, A339278, A262626, A336811, A338156, A340035, A341149, A346533, A350333.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Jan 07 2021