A347367 Irregular triangle read by rows: T(n,k) is the total number of cells with multiplicity in the k-th column of the ziggurat diagram of n.
1, 1, 2, 1, 1, 2, 0, 2, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 3, 0, 0, 0, 3, 2, 1, 1, 2, 3, 4, 5, 7, 5, 4, 3, 2, 1, 1, 2, 3, 4, 0, 0, 0, 0, 0, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 0, 0, 1, 4, 1, 0, 0, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0
Offset: 1
Examples
Triangle begins: 1; 1, 2, 1; 1, 2, 0, 2, 1; 1, 2, 3, 4, 3, 2, 1; 1, 2, 3, 0, 0, 0, 3, 2, 1; 1, 2, 3, 4, 5, 7, 5, 4, 3, 2, 1; 1, 2, 3, 4, 0, 0, 0, 0, 0, 4, 3, 2, 1; 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1; 1, 2, 3, 4, 5, 0, 0, 1, 4, 1, 0, 0, 5, 4, 3, 2, 1; 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 9, 8, 7, 6, 5, 4, 3, 2, 1; 1, 2, 3, 4, 5, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 5, 4, 3, 2, 1; ... Written as an isosceles triangle we can see the symmetry of every row as shown below: 1; 1, 2, 1; 1, 2, 0, 2, 1; 1, 2, 3, 4, 3, 2, 1; 1, 2, 3, 0, 0, 0, 3, 2, 1; 1, 2, 3, 4, 5, 7, 5, 4, 3, 2, 1; 1, 2, 3, 4, 0, 0, 0, 0, 0, 4, 3, 2, 1; 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1; 1, 2, 3, 4, 5, 0, 0, 1, 4, 1, 0, 0, 5, 4, 3, 2, 1; 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 9, 8, 7, 6, 5, 4, 3, 2, 1; 1, 2, 3, 4, 5, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 5, 4, 3, 2, 1; ... For n = 15 the ziggurat diagram of 15 looks like this: _ | | _ | | _ _| | _| |_ | |_ _| | | | | |_ _| | | | | |_ _| | _| |_ | |_ _| | | | | |_ _| | | | | |_ _| | _| _ |_ | |_ |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _| 1 2 3 4 5 6 7 8 0 0 0 1 4 7 B 7 4 1 0 0 0 8 7 6 5 4 3 2 1 . Where B = 10 + 1 = 11. The left-hand part (or the left-hand staircase) has 8 steps. The right-hand part (or the right-hand staircase) has 8 steps. The central part (formed by two subparts or two staircases) has a total of 7 + 1 = 8 steps. The number of parts equals A237271(15) = 3. The number of subparts equals A001227(15) = 4. The number of steps in the central column equals A067742(15) = 2. The total number of steps equals A000203(15) = 24. Compare the above diagram with the symmetric representation of sigma(15) with subparts as shown below: _ | | | | | | | | | | | | | | _ _ _|_| _ _| | 8 | _ _| _| |_| |_ _| 1 | 7 _ _ _ _ _ _ _ _| |_ _ _ _ _ _ _ _| 8 . The left-hand part has 8 square cells. The right-hand part has 8 square cells. The central part (formed by two subparts) has a total of 7 + 1 = 8 square cells. The number of parts equals A237271(15) = 3. The number of subparts equals A001227(15) = 4. The number of square cells on the main diagonal equals A067742(15) = 2. The total number of square cells equals A000203(15) = 24.
Comments