Real-rooted atlas

A257619 → x ↦ x/(x+1)

nonnegative coefficients

Proved for all n all roots real, in [−1, 0]coefficients of one signHadamard kernel

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(2 + 8\,x + 10\,x^{2} + 4\,x^{3}\right)\,P_{n-1}(x) + \left(9\,x + 27\,x^{2} + 27\,x^{3} + 9\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1246
245252108
38588171611443456
4166500419647092835464154872
532715649237802980120354640014185608940456
6647873321940593210674664022714692020852832069509440632124648
712886609083983539563522583064119407288001888670784014110416320403154752052898998536

All 15 rows

Technical details
Node id
n-8dfc00cbda2ca4e8 · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-cc3d28c31b2705bc

Observed on the stored rows ?

  • Degrees: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 (affine: slope 1, intercept 0)
  • Palindromic rows: no
  • γ-positive (gamma_positive): no
  • Nonnegative coefficients (nonn): yes
  • Only rational roots (rr_by_factors: every nonzero row is a product of linear factors over ℚ): no

Matches ?

No OEIS matches.

Submission text ?

Atlas rows n = 0, …, ascending powers of x
%e Example
Triangle begins:
  0:   1;
  1:   2,          4;
  2:   4,         52,           52;
  3:   8,        588,         1716,          1144;
  4:  16,       6500,        41964,         70928,          35464;
  5:  32,      71564,       923780,       2980120,        3546400,        1418560;
  6:  64,     787332,     19405932,     106746640,      227146920,      208528320,        69509440;
  7: 128,    8660908,    398353956,    3522583064,    11940728800,    18886707840,     14110416320,      4031547520;
  8: 256,   95270500,   8079670924,  110918695888,   562947769384,  1365304420480,   1715631998080,   1080454735360,   270113683840;
  9: 512, 1047976524, 162831934980, 3394394941080, 24830494809120, 86687218537920, 162975308496000, 169736213687040, 92378879873280, 20528639971840;
%F Formula
The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (2+8*x+10*x^2+4*x^3)*P(n-1,x) + (9*x+27*x^2+27*x^3+9*x^4)*d/dx P(n-1,x) for n >= 1, with P(0,x) = 1.
%t Mathematica
P[0, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((2 + 8*x + 10*x^2 + 4*x^3)*P[n - 1, x] + (9*x + 27*x^2 + 27*x^3 + 9*x^4)*D[P[n - 1, x], x])/(1 + 2*x + x^2)]];
Table[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N + 1)); v[1] = 1; for(n = 1, N, my(i = n + 1); v[i] = ((2 + 8*x + 10*x^2 + 4*x^3)*v[i - 1] + (9*x + 27*x^2 + 27*x^3 + 9*x^4)*deriv(v[i - 1], x))/(1 + 2*x + x^2)); v};
row(n, p) = {my(c = Vecrev(p)); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};
T(9)

polytool c48629374597

Properties

Proof chains ? JSON

Same fingerprint ?

References

  1. [LW07] Lily L. Liu and Yi Wang. A unified approach to polynomial sequences with only real zeros. Advances in Applied Mathematics, 38(4):542–560, May 2007. doi:10.1016/j.aam.2006.02.003