A162982 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k 3-term arithmetic progressions (n>=0; 0<=k<=floor((n-1)^2/4)).
1, 1, 2, 4, 2, 10, 12, 2, 20, 48, 46, 4, 2, 48, 156, 318, 152, 40, 4, 2, 104, 460, 1112, 1690, 1152, 406, 92, 18, 4, 2, 282, 1248, 4058, 8784, 11648, 8856, 3906, 1188, 244, 80, 20, 4, 2, 496, 2924, 11360, 31776, 64020, 86676, 80700, 52800, 22212, 6948, 2158, 516, 214, 52, 22, 4, 2
Offset: 0
Examples
T(5,3) = 4 because we have 12354 (containing 123, 234, 135), 21345 (containing 234, 345, and 135), and their reversals 45321 and 54312. Triangle starts: 1; 1; 2; 4, 2; 10, 12, 2; 20, 48, 46, 4, 2; 48, 156, 318, 152, 40, 4, 2; ...
Links
- Alois P. Heinz, Rows n = 0..20, flattened
Programs
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Maple
n := 7: with(combinat): P := permute(n): st := proc (p) local ct, i, j, k: ct := 0: for i to nops(p)-2 do for j from i+1 to nops(p)-1 do for k from j+1 to nops(p) do if p[i]+p[k] = 2*p[j] then ct := ct+1 else end if end do end do end do; ct end proc: sort(add(t^st(P[i]), i = 1 .. factorial(n))); # yields the generating polynomial of row n
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Mathematica
row[n_] := CoefficientList[P = Permutations[Range[n]]; st[p_List] := Module[{ct = 0, i, j, k}, For[i = 1, i <= Length[p]-2, i++, For[j = i+1, j <= Length[p]-1, j++, For[k = j+1, k <= Length[p], k++, If[p[[i]] + p[[k]] == 2*p[[j]], ct = ct+1]]]]; ct]; Sum[t^st[P[[i]]], {i, 1, n!}], t]; Table[ro = row[n]; Print[ro]; ro, {n, 0, 9}] // Flatten (* Jean-François Alcover, Sep 08 2017, adapted from Maple *)
Comments