A144439 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(2\,x + 6\,x^{2} + 6\,x^{3} + 2\,x^{4}\right)\,P_{n-2}(x) + \left(1 + 4\,x + 5\,x^{2} + 2\,x^{3}\right)\,P_{n-1}(x) + \left(2\,x + 6\,x^{2} + 6\,x^{3} + 2\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)
\(P_{1}(x) = 1 + 2\,x\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1123
21101021
31349664195
4110663810645322341
5132236841153613620544834611
619701992210434421827219932066440609269
71291410413685795228296404425552329369694105612454947
Technical details
Node id
n-5cd2c82cd24193ca · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-42d134b85075e1e7

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,    10,      10;
  3: 1,    34,      96,       64;
  4: 1,   106,     638,     1064,       532;
  5: 1,   322,    3684,    11536,     13620,       5448;
  6: 1,   970,   19922,   104344,    218272,     199320,      66440;
  7: 1,  2914,  104136,   857952,   2829640,    4425552,    3293696,     941056;
  8: 1,  8746,  534278,  6672264,  32578908,   77622704,   96759856,   60759104,   15189776;
  9: 1, 26242, 2712204, 50125616, 348512460, 1187014008, 2203860960, 2281463040, 1238841360, 275298080;
%F Formula
The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (2*x+6*x^2+6*x^3+2*x^4)*P(n-2,x) + (1+4*x+5*x^2+2*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 >= 2, with P(0,x) = 1, P(1,x) = 1 + 2*x.
%t Mathematica
P[0, x_] := 1;
P[1, x_] := 1 + 2*x;
P[n_, x_] := P[n, x] = Expand[Together[((2*x + 6*x^2 + 6*x^3 + 2*x^4)*P[n - 2, x] + (1 + 4*x + 5*x^2 + 2*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; v[2] = 1 + 2*x; for(n = 2, N, my(i = n + 1); v[i] = ((2*x + 6*x^2 + 6*x^3 + 2*x^4)*v[i - 2] + (1 + 4*x + 5*x^2 + 2*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