%I #10 Jan 29 2024 11:01:39
%S 1,2,2,1,3,4,4,2,1,4,6,7,6,4,2,1,5,8,10,10,9,6,4,2,1,6,10,13,14,14,12,
%T 9,6,4,2,1,7,12,16,18,19,18,16,12,9,6,4,2,1,8,14,19,22,24,24,23,20,16,
%U 12,9,6,4,2,1,9,16,22,26,29,30,30,28,25,20,16
%N Irregular triangular array T, read by rows: T(n,k) = number of sums |x-y| + |y-z| = k, where x,y,z are in {1,2,...,n} and x <= y and y >= z.
%C Row n consists of 2n-1 positive integers.
%e First six rows:
%e 1
%e 2 2 1
%e 3 4 4 2 1
%e 4 6 7 6 4 2 1
%e 5 8 10 10 9 6 4 2 1
%e 6 10 13 14 14 12 9 6 4 2 1
%e For n=2, there are 5 triples (x,y,z) having x <= y and y >= z:
%e 111: |x-y| + |y-z| = 0
%e 121: |x-y| + |y-z| = 2
%e 122: |x-y| + |y-z| = 1
%e 221: |x-y| + |y-z| = 1
%e 222: |x-y| + |y-z| = 0
%e so row 2 of the array is (2,2,1), representing two 0s, two 1s, and one 3.
%t t1[n_] := t1[n] = Tuples[Range[n], 3];
%t t[n_] := t[n] = Select[t1[n], #[[1]] <= #[[2]] >= #[[3]] &];
%t a[n_, k_] := Select[t[n], Abs[#[[1]] - #[[2]]] + Abs[#[[2]] - #[[3]]] == k &];
%t u = Table[Length[a[n, k]], {n, 1, 15}, {k, 0, 2 n - 2}];
%t v = Flatten[u] (* sequence *)
%t Column[Table[Length[a[n, k]], {n, 1, 15}, {k, 0, 2 n - 2}]] (* array *)
%Y Cf. A000027 (column 1), A000330 (row sums), A002620 (limiting reversed row), A368434, A368437, A368515, A368516, A368517, A368518, A368519, A368520, A368521, A368522, A368604, A368605, A368607, A368609.
%K nonn,tabf
%O 1,2
%A _Clark Kimberling_, Jan 22 2024