A034807 → D+1

nonnegative coefficients

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

Recurrence

\(\left(9\,x - 4\,x^{2} + 2\,n - 13\,n x + 4\,n x^{2} - 3\,n^{2} + 4\,n^{2} x + n^{3}\right)\,P_{n}(x) = \left(2 + 11\,x + 6\,x^{2} - 8\,x^{3} - n - 11\,n x - 14\,n x^{2} + 8\,n x^{3} - 2\,n^{2} - 2\,n^{2} x + 8\,n^{2} x^{2} + n^{3} + 2\,n^{3} x\right)\,P_{n-1}(x) + \left(x + 5\,x^{2} - 16\,x^{4} - 4\,n x^{2} - 16\,n x^{3} - n^{2} x - 4\,n^{2} x^{2}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 2\)
\(P_{1}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
022
111
2325
3437
458215
5615526
672415248
783535785
894868242151
91063117669265
101180185150352463
11129927529711011804
12131203905322854821390
1314143533884637169132393
1415168707138612744906324105
Technical details
Node id
n-fcc3d4dfbae91090 · revision 0 · JSON
Definition source
derived:one_plus_derivative
Fingerprint
f6-4d17291db27785d7

Observed on the stored rows ?

  • Degrees: 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7 (half 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:  2;
  1:  1;
  2:  3,  2;
  3:  4,  3;
  4:  5,  8,   2;
  5:  6, 15,   5;
  6:  7, 24,  15,  2;
  7:  8, 35,  35,  7;
  8:  9, 48,  68, 24, 2;
  9: 10, 63, 117, 66, 9;
%F Formula
The row polynomials satisfy (n^3-3*n^2+2*n+(4*n^2-13*n+9)*x+(4*n-4)*x^2)*P(n,x) = (n^3-2*n^2-n+2+(2*n^3-2*n^2-11*n+11)*x+(8*n^2-14*n+6)*x^2+(8*n-8)*x^3)*P(n-1,x) - ((n^2-1)*x+(4*n^2+4*n-5)*x^2+16*n*x^3+16*x^4)*d/dx P(n-1,x) for n >= 2, with P(0,x) = 2, P(1,x) = 1.
%t Mathematica
P[0, x_] := 2;
P[1, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((n^3 - 2*n^2 - n + 2 + (2*n^3 - 2*n^2 - 11*n + 11)*x + (8*n^2 - 14*n + 6)*x^2 + (8*n - 8)*x^3)*P[n - 1, x] - ((n^2 - 1)*x + (4*n^2 + 4*n - 5)*x^2 + 16*n*x^3 + 16*x^4)*D[P[n - 1, x], x])/(n^3 - 3*n^2 + 2*n + (4*n^2 - 13*n + 9)*x + (4*n - 4)*x^2)]];
Table[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N + 1)); v[1] = 2; v[2] = 1; for(n = 2, N, my(i = n + 1); v[i] = ((n^3 - 2*n^2 - n + 2 + (2*n^3 - 2*n^2 - 11*n + 11)*x + (8*n^2 - 14*n + 6)*x^2 + (8*n - 8)*x^3)*v[i - 1] - ((n^2 - 1)*x + (4*n^2 + 4*n - 5)*x^2 + 16*n*x^3 + 16*x^4)*deriv(v[i - 1], x))/(n^3 - 3*n^2 + 2*n + (4*n^2 - 13*n + 9)*x + (4*n - 4)*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
  2. [SP14] J Schur and G Polya. Über zwei Arten von Faktorenfolgen in der Theorie der algebraischen Gleichungen. J. Reine Angew. Math., 1914.