A368521 Triangular array T, read by rows: T(n,k) = number of sums |x-y| + |y-z| - |x-z| = k, where x,y,z are in {1,2,...,n}.
1, 6, 2, 17, 8, 2, 36, 18, 8, 2, 65, 32, 18, 8, 2, 106, 50, 32, 18, 8, 2, 161, 72, 50, 32, 18, 8, 2, 232, 98, 72, 50, 32, 18, 8, 2, 321, 128, 98, 72, 50, 32, 18, 8, 2, 430, 162, 128, 98, 72, 50, 32, 18, 8, 2, 561, 200, 162, 128, 98, 72, 50, 32, 18, 8, 2, 716
Offset: 1
Examples
First eight rows: 1 6 2 17 8 2 36 18 8 2 65 32 18 8 2 106 50 32 18 8 2 161 72 50 32 18 8 2 232 98 72 50 32 18 8 2 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 (6,2), representing six 0s and two 2s.
Crossrefs
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]]] == 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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