A063967 → basis A013611 → D

nonnegative coefficients

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

Recurrence

\(\left(6 + 8\,x - 5\,n - 4\,n x + n^{2}\right)\,P_{n}(x) = \left(-48 - 80\,x - 32\,x^{2} + 22\,n + 20\,n x - 2\,n^{2}\right)\,P_{n-1}(x) + \left(-12 - 44\,x - 48\,x^{2} - 16\,x^{3} + 3\,n + 8\,n x + 4\,n x^{2}\right)\,P_{n-1}'(x)\)
\(P_{1}(x) = 0\)
\(P_{2}(x) = 4\)

Rows ?

nx0x1x2x3x4x5x6Pn(1)
10
244
3203252
480256192528
52881376211210244800
697662081440014336512041040
7318425280785281187848704024576337392
810112962563747847618568499204915201146882699136
Technical details
Node id
n-b21d48540d48c5a7 · revision 0 · JSON
Definition source
derived:derivative
Fingerprint
f6-a004815a7e9d3037

Observed on the stored rows ?

  • Degrees: -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 (affine: slope 1, intercept -2)
  • 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:     0;
   2:     4;
   3:    20,      32;
   4:    80,     256,     192;
   5:   288,    1376,    2112,     1024;
   6:   976,    6208,   14400,    14336,     5120;
   7:  3184,   25280,   78528,   118784,    87040,    24576;
   8: 10112,   96256,  374784,   761856,   849920,   491520,   114688;
   9: 31488,  349440, 1635840,  4178944,  6277120,  5529600,  2637824,   524288;
  10: 96576, 1224192, 6695424, 20604928, 38958080, 46252032, 33603584, 13631488, 2359296;
%F Formula
The row polynomials satisfy (n^2-5*n+6-(4*n-8)*x)*P(n,x) = -(2*n^2-22*n+48-(20*n-80)*x+32*x^2)*P(n-1,x) + (3*n-12+(8*n-44)*x+(4*n-48)*x^2-16*x^3)*d/dx P(n-1,x) for n >= 3, with P(1,x) = 0, P(2,x) = 4.
%t Mathematica
P[1, x_] := 0;
P[2, x_] := 4;
P[n_, x_] := P[n, x] = Expand[Together[(-(2*n^2 - 22*n + 48 - (20*n - 80)*x + 32*x^2)*P[n - 1, x] + (3*n - 12 + (8*n - 44)*x + (4*n - 48)*x^2 - 16*x^3)*D[P[n - 1, x], x])/(n^2 - 5*n + 6 - (4*n - 8)*x)]];
Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N)); v[1] = 0; v[2] = 4; for(n = 3, N, my(i = n); v[i] = (-(2*n^2 - 22*n + 48 - (20*n - 80)*x + 32*x^2)*v[i - 1] + (3*n - 12 + (8*n - 44)*x + (4*n - 48)*x^2 - 16*x^3)*deriv(v[i - 1], x))/(n^2 - 5*n + 6 - (4*n - 8)*x)); 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. [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. [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
  3. [SP14] J Schur and G Polya. Über zwei Arten von Faktorenfolgen in der Theorie der algebraischen Gleichungen. J. Reine Angew. Math., 1914.