A113413 → basis A013609 → x ↦ x+1

nonnegative coefficients

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

Recurrence

\(\left(1 + n\right)\,P_{n}(x) = \left(21 + 8\,x - 13\,n - 4\,n x + 2\,n^{2}\right)\,P_{n-1}(x) + \left(55 + 51\,x + 10\,x^{2} - 14\,n - 12\,n x - 2\,n x^{2}\right)\,P_{n-1}'(x) + \left(21 + 32\,x + 15\,x^{2} + 2\,x^{3}\right)\,P_{n-1}''(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1527
22320447
3107142608317
4497888576160162137
523095186454419364003214407
61072729020320521846458409606497127
74983515771021025215285665360164162240128654797
Technical details
Node id
n-7ca3329c5b61a09d · revision 0 · JSON
Definition source
derived:shift_x_plus_one
Fingerprint
f6-2c26e7d1cc623ede

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,       2;
  2:      23,      20,       4;
  3:     107,     142,      60,       8;
  4:     497,     888,     576,     160,      16;
  5:    2309,    5186,    4544,    1936,     400,      32;
  6:   10727,   29020,   32052,   18464,    5840,     960,     64;
  7:   49835,  157710,  210252,  152856,   65360,   16416,   2240,    128;
  8:  231521,  839024, 1310624, 1151424,  621600,  210944,  43904,   5120,   256;
  9: 1075589, 4391938, 7866720, 8106016, 5291040, 2266944, 637056, 113152, 11520, 512;
%F Formula
The row polynomials satisfy (n+1)*P(n,x) = (2*n^2-13*n+21-(4*n-8)*x)*P(n-1,x) - (14*n-55+(12*n-51)*x+(2*n-10)*x^2)*d/dx P(n-1,x) + (21+32*x+15*x^2+2*x^3)*d^2/dx^2 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[((2*n^2 - 13*n + 21 - (4*n - 8)*x)*P[n - 1, x] - (14*n - 55 + (12*n - 51)*x + (2*n - 10)*x^2)*D[P[n - 1, x], x] + (21 + 32*x + 15*x^2 + 2*x^3)*D[P[n - 1, x], {x, 2}])/(n + 1)]];
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] = ((2*n^2 - 13*n + 21 - (4*n - 8)*x)*v[i - 1] - (14*n - 55 + (12*n - 51)*x + (2*n - 10)*x^2)*deriv(v[i - 1], x) + (21 + 32*x + 15*x^2 + 2*x^3)*deriv(deriv(v[i - 1], x), x))/(n + 1)); 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

Candidate equalities ?

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