A368522 Triangular array T, read by rows: T(n,k) = number of sums |x-y| + |y-z| - |x-z| = 2n-2-k, where x,y,z are in {1,2,...,n}.
1, 2, 6, 2, 8, 17, 2, 8, 18, 36, 2, 8, 18, 32, 65, 2, 8, 18, 32, 50, 106, 2, 8, 18, 32, 50, 72, 161, 2, 8, 18, 32, 50, 72, 98, 232, 2, 8, 18, 32, 50, 72, 98, 128, 321, 2, 8, 18, 32, 50, 72, 98, 128, 162, 430, 2, 8, 18, 32, 50, 72, 98, 128, 162, 200, 561, 2
Offset: 1
Examples
First eight rows: 1 2 6 2 8 17 2 8 18 36 2 8 18 32 65 2 8 18 32 50 106 2 8 18 32 50 72 161 2 8 18 32 50 72 98 232 For n=2, there are 8 triples (x,y,z): 111: |x-y| + |y-z| - |x-z| = 0 112: |x-y| + |y-z| - |x-z| = 0 121: |x-y| + |y-z| - |x-z| = 2 122: |x-y| + |y-z| - |x-z| = 0 211: |x-y| + |y-z| - |x-z| = 0 212: |x-y| + |y-z| - |x-z| = 2 221: |x-y| + |y-z| - |x-z| = 0 222: |x-y| + |y-z| - |x-z| = 0 so row 2 of the array is (2,6), representing two 2s and six 0s.
Programs
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Mathematica
t[n_] := t[n] = Tuples[Range[n], 3] a[n_, k_] := Select[t[n], Abs[#[[1]] - #[[2]]] + Abs[#[[2]] - #[[3]]] - Abs[#[[1]] - #[[3]]] == 2n-2-k &] u = Table[Length[a[n, k]], {n, 1, 15}, {k, 0, 2 n - 2, 2}] v = Flatten[u] (* sequence *) Column[Table[Length[a[n, k]], {n, 1, 15}, {k, 0, 2 n - 2, 2}]] (* array *)
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