A257613 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(4 + 16\,x + 20\,x^{2} + 8\,x^{3}\right)\,P_{n-1}(x) + \left(2\,x + 6\,x^{2} + 6\,x^{3} + 2\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
14812
2168080176
36460814409603072
4256416017600268801344062336
51024270081824004800005376002150401443072
64096170240172928069888001317120011612160387072037546496
71638410542081553664090639360255897600373309440270950400774144001084818432
Technical details
Node id
n-96cfdd1a383cbf1a · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-cef044b741d89456

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:      4,        8;
  2:     16,       80,         80;
  3:     64,      608,       1440,         960;
  4:    256,     4160,      17600,       26880,       13440;
  5:   1024,    27008,     182400,      480000,      537600,       215040;
  6:   4096,   170240,    1729280,     6988800,    13171200,     11612160,      3870720;
  7:  16384,  1054208,   15536640,    90639360,   255897600,    373309440,    270950400,     77414400;
  8:  65536,  6456320,  134835200,  1092833280,  4339722240,   9320693760,  11054776320,   6812467200,   1703116800;
  9: 262144, 39262208, 1143244800, 12546355200, 67376332800, 199925268480, 344648908800, 343719936000, 183936614400, 40874803200;
%F Formula
The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (4+16*x+20*x^2+8*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[((4 + 16*x + 20*x^2 + 8*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] = ((4 + 16*x + 20*x^2 + 8*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