Real-rooted atlas

A257624 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(5 + 20\,x + 25\,x^{2} + 10\,x^{3}\right)\,P_{n-1}(x) + \left(3\,x + 9\,x^{2} + 9\,x^{3} + 3\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
151015
225130130285
31251290312020806615
462511570510907904039520181845
5312598810712400192400021736008694405781375
615625821730912093038334400735072006520800021736000208743885
77812567300901110127206826164801977976000292131840021301280006086080008438467815

All 15 rows

Technical details
Node id
n-96937d0c31038dc0 · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-1d32ce7fdc01b2c4

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:       5,        10;
  2:      25,       130,         130;
  3:     125,      1290,        3120,         2080;
  4:     625,     11570,       51090,        79040,         39520;
  5:    3125,     98810,      712400,      1924000,       2173600,        869440;
  6:   15625,    821730,     9120930,     38334400,      73507200,      65208000,      21736000;
  7:   78125,   6730090,   111012720,    682616480,    1977976000,    2921318400,    2130128000,     608608000;
  8:  390625,  54621970,  1308631090,  11332834240,   46595305120,  101941840000,  122025904000,   75467392000,   18866848000;
  9: 1953125, 440882010, 15105027600, 179597776800, 1007444037600, 3063933512640, 5355141792000, 5378877504000, 2886627744000, 641472832000;
%F Formula
The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (5+20*x+25*x^2+10*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[((5 + 20*x + 25*x^2 + 10*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] = ((5 + 20*x + 25*x^2 + 10*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