Real-rooted atlas

A257609 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(2 + 8\,x + 10\,x^{2} + 4\,x^{3}\right)\,P_{n-1}(x) + \left(2\,x + 6\,x^{2} + 6\,x^{3} + 2\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1246
24242452
38112288192600
4164802400384019208656
532198417280499205760023040149856
664806411558453760010752009676803225603026752
7128325127418885225472161280002451456018063360516096069866880

All 15 rows

Technical details
Node id
n-9d90ef1d46f325d6 · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-a83eeccfb5efd735

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,     24,       24;
  3:   8,    112,      288,       192;
  4:  16,    480,     2400,      3840,       1920;
  5:  32,   1984,    17280,     49920,      57600,      23040;
  6:  64,   8064,   115584,    537600,    1075200,     967680,      322560;
  7: 128,  32512,   741888,   5225472,   16128000,   24514560,    18063360,     5160960;
  8: 256, 130560,  4646400,  47738880,  213534720,  487710720,   596090880,   371589120,   92897280;
  9: 512, 523264, 28661760, 419082240, 2612736000, 8414945280, 15173222400, 15482880000, 8360755200, 1857945600;
%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) + (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[((2 + 8*x + 10*x^2 + 4*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] = ((2 + 8*x + 10*x^2 + 4*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