Real-rooted atlas

A112007 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

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

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1235
26201541
324130210105469
4120924238025209456889
5720730826432441003465010395123605
65040642243036607053208662505405401351352620169
74032062337636788401109878018858840182882709459450202702564074901

All 15 rows

Technical details
Node id
n-e977d16b43a5fb43 · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-0c4bf0a352e1721b

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,        3;
  2:       6,       20,        15;
  3:      24,      130,       210,        105;
  4:     120,      924,      2380,       2520,        945;
  5:     720,     7308,     26432,      44100,      34650,       10395;
  6:    5040,    64224,    303660,     705320,     866250,      540540,      135135;
  7:   40320,   623376,   3678840,   11098780,   18858840,    18288270,     9459450,     2027025;
  8:  362880,  6636960,  47324376,  177331440,  389449060,   520059540,   416215800,   183783600,   34459425;
  9: 3628800, 76998240, 647536032, 2920525608, 7934927000, 13642629000, 14980405440, 10199989800, 3928374450, 654729075;
%F Formula
The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (n+1+(3*n+4)*x+(3*n+5)*x^2+(n+2)*x^3)*P(n-1,x) + (x+3*x^2+3*x^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[((n + 1 + (3*n + 4)*x + (3*n + 5)*x^2 + (n + 2)*x^3)*P[n - 1, x] + (x + 3*x^2 + 3*x^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] = ((n + 1 + (3*n + 4)*x + (3*n + 5)*x^2 + (n + 2)*x^3)*v[i - 1] + (x + 3*x^2 + 3*x^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