A019538 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(1 + 7\,x + 15\,x^{2} + 13\,x^{3} + 4\,x^{4} - n x - 4\,n x^{2} - 5\,n x^{3} - 2\,n x^{4}\right)\,P_{n-1}(x) + \left(x + 5\,x^{2} + 9\,x^{3} + 7\,x^{4} + 2\,x^{5}\right)\,P_{n-1}'(x)\)
\(P_{1}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
111
2134
3181322
41176775160
51342466345411456
616779835626793468315904
71132245116904541118178047293202672
8126173417364935237187625510959675458352951680
Technical details
Node id
n-6042ba4cd721015a · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-6c43fbf0495b302d

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 -1)
  • 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 = 1, …, ascending powers of x
%e Example
Triangle begins:
   1: 1;
   2: 1,    3;
   3: 1,    8,    13;
   4: 1,   17,    67,      75;
   5: 1,   34,   246,     634,      541;
   6: 1,   67,   798,    3562,     6793,     4683;
   7: 1,  132,  2451,   16904,    54111,    81780,     47293;
   8: 1,  261,  7341,   73649,   352371,   876255,   1095967,    545835;
   9: 1,  518, 21748,  306146,  2056690,  7487306,  15196348,  16205198,   7087261;
  10: 1, 1031, 64192, 1239080, 11247058, 56259206, 164793976, 282270272, 262283125, 102247563;
%F Formula
The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (1-(n-7)*x-(4*n-15)*x^2-(5*n-13)*x^3-(2*n-4)*x^4)*P(n-1,x) + (x+5*x^2+9*x^3+7*x^4+2*x^5)*d/dx P(n-1,x) for n >= 2, with P(1,x) = 1.
%t Mathematica
P[1, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((1 - (n - 7)*x - (4*n - 15)*x^2 - (5*n - 13)*x^3 - (2*n - 4)*x^4)*P[n - 1, x] + (x + 5*x^2 + 9*x^3 + 7*x^4 + 2*x^5)*D[P[n - 1, x], x])/(1 + 2*x + x^2)]];
Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N)); v[1] = 1; for(n = 2, N, my(i = n); v[i] = ((1 - (n - 7)*x - (4*n - 15)*x^2 - (5*n - 13)*x^3 - (2*n - 4)*x^4)*v[i - 1] + (x + 5*x^2 + 9*x^3 + 7*x^4 + 2*x^5)*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, v[i])))};
T(10)

polytool c48629374597

Properties

Proof chains ? JSON

Same fingerprint ?

References

  1. [MW08] Shi-Mei Ma and Yi Wang. q-Eulerian polynomials and polynomials with only real zeros. The Electronic Journal of Combinatorics, 15(1):Research Paper 17, January 2008. doi:10.37236/741