Real-rooted atlas

A257611 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(3 + 12\,x + 15\,x^{2} + 6\,x^{3}\right)\,P_{n-1}(x) + \left(2\,x + 6\,x^{2} + 6\,x^{3} + 2\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1369
294848105
3272947204801521
4811632739211520576026385
524386466480017760020160080640533529
672944688522768224640043488003870720129024012324345
72187227814401688025445280747936001112025608128512023224320320197761

All 15 rows

Technical details
Node id
n-ad6ee8efc8bbdf6b · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-f526dd4eeb36ec75

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:     3,       6;
  2:     9,      48,        48;
  3:    27,     294,       720,        480;
  4:    81,    1632,      7392,      11520,        5760;
  5:   243,    8646,     64800,     177600,      201600,       80640;
  6:   729,   44688,    522768,    2246400,     4348800,     3870720,     1290240;
  7:  2187,  227814,   4016880,   25445280,    74793600,   111202560,    81285120,    23224320;
  8:  6561, 1152192,  29940672,  269176320,  1128072960,  2492743680,  2998517760,  1857945600,   464486400;
  9: 19683, 5800326, 218802240, 2721993600, 15638918400, 48198689280, 84861665280, 85558394880, 45984153600, 10218700800;
%F Formula
The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (3+12*x+15*x^2+6*x^3)*P(n-1,x) + (2*x+6*x^2+6*x^3+2*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[((3 + 12*x + 15*x^2 + 6*x^3)*P[n - 1, x] + (2*x + 6*x^2 + 6*x^3 + 2*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] = ((3 + 12*x + 15*x^2 + 6*x^3)*v[i - 1] + (2*x + 6*x^2 + 6*x^3 + 2*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