A380385 - OEIS

%I #20 Jan 25 2025 13:54:07

%S 1,0,1,0,-1,1,0,0,-1,1,0,0,-1,-1,1,0,0,1,-1,-1,1,0,0,0,0,-1,-1,1,0,0,

%T 0,1,0,-1,-1,1,0,0,0,1,0,0,-1,-1,1,0,0,0,-1,2,0,0,-1,-1,1,0,0,0,0,0,1,

%U 0,0,-1,-1,1,0,0,0,0,0,1,1,0,0,-1,-1,1

%N Triangle read by rows. T(1, 1) = 1, T(n, k) = [n >= k](Sum_{i=1..k-1} T[n - i, k - 1] - Sum_{i=1..n-1} T[n - i, k]).

%C For n >= 1 the row sum of row n appears to be -A010815(n). See the Mathematica program for exact formulation of the recurrence.

%e {

%e {1},

%e {0,  1},

%e {0, -1,  1},

%e {0,  0, -1,  1},

%e {0,  0, -1, -1,  1},

%e {0,  0,  1, -1, -1,  1},

%e {0,  0,  0,  0, -1, -1,  1},

%e {0,  0,  0,  1,  0, -1, -1,  1},

%e {0,  0,  0,  1,  0,  0, -1, -1,  1},

%e {0,  0,  0, -1,  2,  0,  0, -1, -1,  1},

%e {0,  0,  0,  0,  0,  1,  0,  0, -1, -1,  1},

%e {0,  0,  0,  0,  0,  1,  1,  0,  0, -1, -1,  1}

%e }

%t nn = 12; t[n_, 1] = If[n == 1, 1, 0]; t[n_, k_] := t[n, k] = If[n >= k, Sum[t[n - i, k - 1], {i, 1, k - 1}] - Sum[t[n - i, k], {i, 1, n - 1}], 0]; Table[Table[t[n, k], {k, 1, n}], {n, 1, nn}]; Flatten[%]

%t nn = 12; Table[Table[If[n == k, 1, Coefficient[Product[(1 - x^i), {i, k - 1}], x^(n - k)]], {k, 1, n}], {n, 1, nn}] // Flatten (* Conjectured after A231599. - _Mats Granvik_, Jan 25 2025 *)

%Y Cf. A231599, A010815, A008284.

%K sign,tabl

%O 1,50

%A _Mats Granvik_, Jan 23 2025