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Craven and G. Csordas, Multiplier sequences for fields, Illinois J. Math. 21 (1977); Polya--Schur, Theorem 40 of P. Branden, arXiv:1410.6601: sum_k f(k) x^k / k! = f(theta) e^x = C e^x x^m prod (1 + x/a_j), a_j > 0); f(k) has one sign for k >= 0, so coefficients of one sign stay of one sign; applied to f(n, .) for each n","lean_route":null,"lean_declarations":["RealRooted.isMultiplierSequence_eval_of_roots_nonpos","RealRooted.isPFMultiplierSequence_eval_of_roots_nonpos","RealRooted.isMultiplierSequence_iff_preserves"],"certificate_hash":"7b7a5053828714c61064960aa8ca3a9e9e8124363f7cb6280cde11687e8864a5","verification_polytool":"replayed","verification_lean":"not_attempted","claims":[{"claim":"real_rooted","from_index":1,"region":{"upper":{"closed":true,"value":"0"}}},{"claim":"interlacing","from_index":2,"kind":"weak","orientation":"unordered"}],"premises":[{"position":0,"node_id":"n-9a3eef8528def021","role":"sequence","name":"Triangle T(n, k) = coefficients of ( t(n, x) ) where t(n, x) = (1-x)^(n+1)*p(n, x)/x, p(n, x) = x*D( p(n-1, x) ), with p(1, x) = x/(1-x)^2, p(2, x) = x*(1+x)/(1-x)^3, and p(3, x) = x*(1+6*x+x^2)/(1-x)^4, read by rows.","uses":{"claim":"real_rooted","from_index":1,"region":{"upper":{"closed":true,"value":"0"}}},"uses_text":"Every row P_n, n >= 1, 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":1,"oeis_offset":1,"reversed":false,"trim_zeros":"none","regular_triangle":false,"rows_shown":10,"polytool_commit":"c48629374597092cc41ef4d6273f37991516f984","triangle_example":"Triangle begins:\n   1: 1,    24,       9;\n   2: 1,    60,     135,        16;\n   3: 1,   136,     810,       544,         25;\n   4: 1,   292,    3654,      6496,       1825,         36;\n   5: 1,   608,   14247,     51968,      39575,       5472,          49;\n   6: 1,  1244,   50949,    334512,     522675,     203796,       15239,         64;\n   7: 1,  2520,  172440,   1881504,    5226750,    4233384,      938840,      40320,         81;\n   8: 1,  5076,  562680,   9671936,   43772850,   63032904,    29620304,    4001280,     102789,      100;\n   9: 1, 10192, 1790829,  46690304,  324651050,  756394848,   636316058,  186761216,   16117461,   254800,    121;\n  10: 1, 20428, 5601807, 215414480, 2206884050, 7810851672, 10631436998, 5649623168, 1090535805, 62242300, 617947, 144;\n","recurrence_sentence":"The row polynomials satisfy (n+1)*P(n,x) = -((4*n^2+8*n)*x-(9*n^3+25*n^2+16*n)*x^2)*P(n-2,x) - ((2*n+4)*x-(4*n^2+16*n+17)*x^2-(4*n^3-2*n^2-29*n-23)*x^3)*d/dx P(n-2,x) - ((n+2)*x^2-(4*n^2+12*n+9)*x^3+(4*n^2+11*n+7)*x^4)*d^2/dx^2 P(n-2,x) + (n+1+(4*n^2+8*n+4)*x)*P(n-1,x) + ((2*n+3)*x-(4*n^2+12*n+8)*x^2)*d/dx P(n-1,x) for n >= 3, with P(1,x) = 1 + 24*x + 9*x^2, P(2,x) = 1 + 60*x + 135*x^2 + 16*x^3.","mathematica":"P[1, x_] := 1 + 24*x + 9*x^2;\nP[2, x_] := 1 + 60*x + 135*x^2 + 16*x^3;\nP[n_, x_] := P[n, x] = Expand[Together[(-((4*n^2 + 8*n)*x - (9*n^3 + 25*n^2 + 16*n)*x^2)*P[n - 2, x] - ((2*n + 4)*x - (4*n^2 + 16*n + 17)*x^2 - (4*n^3 - 2*n^2 - 29*n - 23)*x^3)*D[P[n - 2, x], x] - ((n + 2)*x^2 - (4*n^2 + 12*n + 9)*x^3 + (4*n^2 + 11*n + 7)*x^4)*D[P[n - 2, x], {x, 2}] + (n + 1 + (4*n^2 + 8*n + 4)*x)*P[n - 1, x] + ((2*n + 3)*x - (4*n^2 + 12*n + 8)*x^2)*D[P[n - 1, x], x])/(n + 1)]];\nTable[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten","pari":"P(N) = {my(v = vector(N)); v[1] = 1 + 24*x + 9*x^2; v[2] = 1 + 60*x + 135*x^2 + 16*x^3; for(n = 3, N, my(i = n); v[i] = (-((4*n^2 + 8*n)*x - (9*n^3 + 25*n^2 + 16*n)*x^2)*v[i - 2] - ((2*n + 4)*x - (4*n^2 + 16*n + 17)*x^2 - (4*n^3 - 2*n^2 - 29*n - 23)*x^3)*deriv(v[i - 2], x) - ((n + 2)*x^2 - (4*n^2 + 12*n + 9)*x^3 + (4*n^2 + 11*n + 7)*x^4)*deriv(deriv(v[i - 2], x), x) + (n + 1 + (4*n^2 + 8*n + 4)*x)*v[i - 1] + ((2*n + 3)*x - (4*n^2 + 12*n + 8)*x^2)*deriv(v[i - 1], x))/(n + 1)); v};\nrow(n, p) = {my(c = Vecrev(p)); c};\nT(N) = {my(v = P(N)); concat(vector(#v, i, row(i, v[i])))};\nT(10)","reading":{"first_row":1,"offset":1,"reversed":false,"trim_zeros":"none","regular_triangle":false}}],"derivation":{"root":"n-9a3eef8528def021","steps":["(\u03b8+1)\u00b2"],"edges":["e-7d11e4cbcf8d0d60"],"text":"A166344 \u2192 (\u03b8+1)\u00b2"},"headline":"A166344 \u2192 (\u03b8+1)\u00b2","chain":null}