Real-rooted atlas

A118931 → basis A013609 → x ↦ x+1

nonnegative coefficients

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

Recurrence

\(P_{n}(x) = \left(6 + 4\,x - 9\,n - 6\,n x + 3\,n^{2} + 2\,n^{2} x\right)\,P_{n-3}(x) + \left(1\right)\,P_{n-1}(x)\)
\(P_{0}(x) = 1\)
\(P_{1}(x) = 1\)
\(P_{2}(x) = 1\)

Rows ?

nx0x1x2Pn(1)
011
111
211
37411
4251641
56140101
64815601601201
72731350011207351

All 15 rows

Technical details
Node id
n-fbfc2c564a29cf31 · revision 0 · JSON
Definition source
derived:shift_x_plus_one
Fingerprint
f6-1c8baaf2c66c380d

Observed on the stored rows ?

  • Degrees: 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4 (irregular)
  • 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:     1;
  3:     7,      4;
  4:    25,     16;
  5:    61,     40;
  6:   481,    560,   160;
  7:  2731,   3500,  1120;
  8: 10417,  13664,  4480;
  9: 91225, 161616, 94080, 17920;
%F Formula
The row polynomials satisfy P(n,x) = (3*n^2-9*n+6+(2*n^2-6*n+4)*x)*P(n-3,x) + P(n-1,x) for n >= 3, with P(0,x) = 1, P(1,x) = 1, P(2,x) = 1.
%t Mathematica
P[0, x_] := 1;
P[1, x_] := 1;
P[2, x_] := 1;
P[n_, x_] := P[n, x] = Expand[(3*n^2 - 9*n + 6 + (2*n^2 - 6*n + 4)*x)*P[n - 3, x] + P[n - 1, x]];
Table[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N + 1)); v[1] = 1; v[2] = 1; v[3] = 1; for(n = 3, N, my(i = n + 1); v[i] = (3*n^2 - 9*n + 6 + (2*n^2 - 6*n + 4)*x)*v[i - 3] + v[i - 1]); 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