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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)).
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%I #23 Nov 02 2021 21:26:47

%S 1,1,2,4,2,10,12,2,20,48,46,4,2,48,156,318,152,40,4,2,104,460,1112,

%T 1690,1152,406,92,18,4,2,282,1248,4058,8784,11648,8856,3906,1188,244,

%U 80,20,4,2,496,2924,11360,31776,64020,86676,80700,52800,22212,6948,2158,516,214,52,22,4,2

%N 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)).

%C Row n contains 1+floor((n-1)^2/4) entries.

%C Sum of entries in row n = n! = A000142(n).

%C T(n,0) = A003407(n).

%C The terms of the sequence have been determined by direct counting (using Maple).

%C The Maple program yields the generating polynomial of the specified row n.

%H Alois P. Heinz, <a href="/A162982/b162982.txt">Rows n = 0..20, flattened</a>

%e T(5,3) = 4 because we have 12354 (containing 123, 234, 135), 21345 (containing 234, 345, and 135), and their reversals 45321 and 54312.

%e Triangle starts:

%e 1;

%e 1;

%e 2;

%e 4, 2;

%e 10, 12, 2;

%e 20, 48, 46, 4, 2;

%e 48, 156, 318, 152, 40, 4, 2;

%e ...

%p 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

%t 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];

%t Table[ro = row[n]; Print[ro]; ro, {n, 0, 9}] // Flatten (* _Jean-François Alcover_, Sep 08 2017, adapted from Maple *)

%Y Cf. A000142, A003407, A295390.

%K nonn,tabf

%O 0,3

%A _Emeric Deutsch_, Aug 31 2009