Real-rooted atlas

A142460 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(1 + 4\,x + 5\,x^{2} + 2\,x^{3}\right)\,P_{n-1}(x) + \left(5\,x + 15\,x^{2} + 15\,x^{3} + 5\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1123
21141429
3186252168507
4151833745712285612461
5131104074013188015708062832395643
6118662469910259896055406405089392169646415414029
7111197452996444722228016053576025421827219000396854286848711678747

All 15 rows

Technical details
Node id
n-861b0207a722f305 · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-b94fba28f209154b

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: 1,       2;
  2: 1,      14,        14;
  3: 1,      86,       252,         168;
  4: 1,     518,      3374,        5712,         2856;
  5: 1,    3110,     40740,      131880,       157080,        62832;
  6: 1,   18662,    469910,     2598960,      5540640,      5089392,      1696464;
  7: 1,  111974,   5299644,    47222280,    160535760,    254218272,    190003968,     54286848;
  8: 1,  671846,  59079902,   819152208,   4174029720,  10141461792,  12754016352,   8034453504,   2008613376;
  9: 1, 4031078, 654581844, 13815394152, 101580211656, 355506660432, 669193975296, 697368849408, 379627928064, 84361761792;
%F Formula
The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (1+4*x+5*x^2+2*x^3)*P(n-1,x) + (5*x+15*x^2+15*x^3+5*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[((1 + 4*x + 5*x^2 + 2*x^3)*P[n - 1, x] + (5*x + 15*x^2 + 15*x^3 + 5*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] = ((1 + 4*x + 5*x^2 + 2*x^3)*v[i - 1] + (5*x + 15*x^2 + 15*x^3 + 5*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