Real-rooted atlas

A257610 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(2 + 8\,x + 10\,x^{2} + 4\,x^{3}\right)\,P_{n-1}(x) + \left(3\,x + 9\,x^{2} + 9\,x^{3} + 3\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1246
24282860
38156420280864
41681244527280364016200
53241244130012460014560058240373896
66420748359268178360036582003319680110656010248120
7128103996301938023212280744016001149657608520512024344320325252584

All 15 rows

Technical details
Node id
n-639fa66408a02be6 · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-dbcd952a44050fb4

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,      28,        28;
  3:   8,     156,       420,        280;
  4:  16,     812,      4452,       7280,        3640;
  5:  32,    4124,     41300,     124600,      145600,       58240;
  6:  64,   20748,    359268,    1783600,     3658200,     3319680,      1106560;
  7: 128,  103996,   3019380,   23212280,    74401600,   114965760,     85205120,     24344320;
  8: 256,  520492,  24883012,  285528880,  1343382040,  3144843520,   3888451840,   2434432000,   608608000;
  9: 512, 2603484, 202707540, 3389647800, 22519224000, 74956452480, 137521063680, 141537876480, 76684608000, 17041024000;
%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) + (3*x+9*x^2+9*x^3+3*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] + (3*x + 9*x^2 + 9*x^3 + 3*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] + (3*x + 9*x^2 + 9*x^3 + 3*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