A368605 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.
1, 1, 2, 3, 2, 1, 3, 5, 5, 4, 2, 1, 4, 7, 8, 8, 6, 4, 2, 1, 5, 9, 11, 12, 11, 9, 6, 4, 2, 1, 6, 11, 14, 16, 16, 15, 12, 9, 6, 4, 2, 1, 7, 13, 17, 20, 21, 21, 19, 16, 12, 9, 6, 4, 2, 1, 8, 15, 20, 24, 26, 27, 26, 24, 20, 16, 12, 9, 6, 4, 2, 1, 9, 17, 23, 28
Offset: 1
Examples
First six rows: 1 1 2 3 2 1 3 5 5 4 2 1 4 7 8 8 6 4 2 1 5 9 11 12 11 9 6 4 2 1 6 11 14 16 16 15 12 9 6 4 2 1 For n=3, there are 8 triples (x,y,z) having x < y and y >= z: 121: |x-y| + |y-z| = 2 122: |x-y| + |y-z| = 1 131: |x-y| + |y-z| = 4 132: |x-y| + |y-z| = 3 133: |x-y| + |y-z| = 2 231: |x-y| + |y-z| = 3 232: |x-y| + |y-z| = 2 233: |x-y| + |y-z| = 1 so row 1 of the array is (2,3,2,1), representing two 1s, three 2s, two 3s, and one 4.
Crossrefs
Programs
-
Mathematica
t1[n_] := t1[n] = Tuples[Range[n], 3]; t[n_] := t[n] = Select[t1[n], #[[1]] < #[[2]] >= #[[3]] &]; a[n_, k_] := Select[t[n], Abs[#[[1]] - #[[2]]] + Abs[#[[2]] - #[[3]]] == k &]; u = Table[Length[a[n, k]], {n, 2, 15}, {k, 1, 2 n - 2}]; v = Flatten[u] (* sequence *) Column[Table[Length[a[n, k]], {n, 2, 15}, {k, 1, 2 n - 2}]] (* array *)
Comments