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v[1] = 1; for(n = 1, N, my(i = n + 1); v[i] = (n - 2 + x)*v[i - 1]); v};\nrow(n, p) = {my(c = Vecrev(p)); c};\nT(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};\nT(9)","reading":{"first_row":0,"offset":0,"reversed":false,"trim_zeros":"none","regular_triangle":false}},{"oeis_id":"A094645","target":"triangle","first_row":0,"oeis_offset":0,"reversed":false,"trim_zeros":"none","regular_triangle":true,"rows_shown":10,"polytool_commit":"c48629374597092cc41ef4d6273f37991516f984","triangle_example":"Triangle begins:\n  0:  1;\n  1: -1,     1;\n  2:  0,    -1,     1;\n  3:  0,    -1,     0,   1;\n  4:  0,    -2,    -1,   2,    1;\n  5:  0,    -6,    -5,   5,    5,    1;\n  6:  0,   -24,   -26,  15,   25,    9,    1;\n  7:  0,  -120,  -154,  49,  140,   70,   14,   1;\n  8:  0,  -720, -1044, 140,  889,  560,  154,  20,  1;\n  9:  0, -5040, -8028, -64, 6363, 4809, 1638, 294, 27, 1;\n","recurrence_sentence":"The row polynomials satisfy P(n,x) = (n-2+x)*P(n-1,x) for n >= 1, with P(0,x) = 1.","mathematica":"P[0, x_] := 1;\nP[n_, x_] := P[n, x] = Expand[(n - 2 + x)*P[n - 1, x]];\nTable[PadRight[CoefficientList[P[n, x], x], n + 1], {n, 0, 9}] // Flatten","pari":"P(N) = {my(v = vector(N + 1)); v[1] = 1; for(n = 1, N, my(i = n + 1); v[i] = (n - 2 + x)*v[i - 1]); v};\nrow(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n + 1 - #c)); c};\nT(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};\nT(9)","reading":{"first_row":0,"offset":0,"reversed":false,"trim_zeros":"none","regular_triangle":true}}],"derivation":{"root":"n-4f6e747b794bcc2e","steps":["x \u21a6 2x\u22123"],"edges":["e-0658035b499d17ce"],"text":"A028338 \u2192 x \u21a6 2x\u22123"},"headline":"A094645: Triangle of generalized Stirling numbers of the first kind read by rows: T(n, k) = (-1)^(n+k)*Sum_{m=0..n} Stirling1(n, m) * binomial(m, k), where Stirling1 is A048994.","chain":"A028338 \u2192 x \u21a6 2x\u22123"}