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the step is Ma and Wang, EJC 15 (2008) R17, Theorem 2)","lean_route":null,"lean_declarations":[],"certificate_hash":"50cc4f7f8c5662751376ed65a3249e6a92163880a2ed34f6edf97ef66db4dfd4","verification_polytool":"replayed","verification_lean":"not_attempted","claims":[{"claim":"real_rooted","from_index":0,"region":{"upper":{"closed":true,"value":"0"}}}],"premises":[{"position":0,"node_id":"n-a44eea8c1f4cf4e5","role":"sequence","name":"Triangular T(n,k) = T(n-1, k) + T(n-1, k-1) + 5*T(n-2, k-1), read by rows.","uses":{"claim":"real_rooted","from_index":0,"region":{"upper":{"closed":true,"value":"0"}}},"uses_text":"Every row P_n, n >= 0, is zero or real-rooted with all roots in (-inf, 0]"}]}],"used_by":[],"used_by_total":0,"links":[],"oeis":[],"bfiles":[],"oeis_texts":[{"oeis_id":"","target":"node","first_row":0,"oeis_offset":0,"reversed":false,"trim_zeros":"none","regular_triangle":false,"rows_shown":10,"polytool_commit":"c48629374597092cc41ef4d6273f37991516f984","triangle_example":"Triangle begins:\n  0:      2;\n  1:     10,        5;\n  2:     50,       52,        2;\n  3:    250,      506,      135,        2;\n  4:    750,     2538,     1336,      155,        2;\n  5:   2750,    14674,    12437,     2826,      177,       2;\n  6:   9250,    75902,    97152,    36347,     4713,     201,      2;\n  7:  32250,   397446,   724435,   397592,    83180,    7065,    227,     2;\n  8: 110750,  2029858,  5131976,  3925095,  1196232,  162230,   9956,   255,   2;\n  9: 382750, 10336034, 35411517, 36331466, 15169947, 2959692, 285148, 13466, 285, 2;\n","recurrence_sentence":"The row polynomials satisfy (n-3)*P(n,x) = (5*n-15+(5*n-15)*x)*P(n-2,x) + (5*n-15)*x*d/dx P(n-2,x) + (2*n-6+(n-3)*x)*P(n-1,x) + (n-3)*x*d/dx P(n-1,x) for n >= 4, with P(0,x) = 2, P(1,x) = 10 + 5*x, P(2,x) = 50 + 52*x + 2*x^2, P(3,x) = 250 + 506*x + 135*x^2 + 2*x^3.","mathematica":"P[0, x_] := 2;\nP[1, x_] := 10 + 5*x;\nP[2, x_] := 50 + 52*x + 2*x^2;\nP[3, x_] := 250 + 506*x + 135*x^2 + 2*x^3;\nP[n_, x_] := P[n, x] = Expand[Together[((5*n - 15 + (5*n - 15)*x)*P[n - 2, x] + (5*n - 15)*x*D[P[n - 2, x], x] + (2*n - 6 + (n - 3)*x)*P[n - 1, x] + (n - 3)*x*D[P[n - 1, x], x])/(n - 3)]];\nTable[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten","pari":"P(N) = {my(v = vector(N + 1)); 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