A249250 → reverse

nonnegative coefficients

Proved for all n: all roots real, in (−∞, 0] · coefficients of one sign · Hadamard kernel what this means

Recurrence
\(\left(-1 + n\right)\,P_{n}(x) = \left(-1 + 4\,x + 2\,x^{2} + n - 7\,n x - 2\,n x^{2} + 3\,n^{2} x\right)\,P_{n-1}(x) + \left(x + 25\,x^{2} + 4\,x^{3} - n x - 10\,n x^{2} - n x^{3}\right)\,P_{n-1}'(x) + \left(x^{2} + 10\,x^{3} + x^{4}\right)\,P_{n-1}''(x)\)
\(P_{0}(x) = 1\)
\(P_{1}(x) = 3 + x\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1314
236110
33159128
43243612176
53339066151208
6342171228105181568
73512795794651532111552

Generated b-files: A249250 (reversed triangle) ?

Computed in your browser ?

Technical details
Node id
n-3cba0f65e01f06f3 · revision 1 · JSON
Definition source
derived:reversal
Fingerprint
f6-034bc45f51d0b25c

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

OEIS matches ?

A generated b-file is computed from the node's definition along the row alignment of its OEIS link and starts at the entry's offset. It is an identification on a finite prefix, not a proof that the node and the OEIS entry agree for every n.

OEIS submission text ?

Fields to paste into an OEIS submission: the example %e, the formula %F and programs %t (Mathematica) and %o (PARI/GP) that list the same rows. For an OEIS link the row numbers, the offset and the orientation are those of the entry, as matched; the matched terms were compared with the published ones, which identifies the node with the entry on a finite prefix, not for every n. The PARI/GP program is not machine-checked here.

The atlas's own rows: n = 0, …, ascending powers of x
%e Example
Triangle begins:
  0: 1;
  1: 3,  1;
  2: 3,  6,   1;
  3: 3, 15,   9,    1;
  4: 3, 24,  36,   12,    1;
  5: 3, 33,  90,   66,   15,    1;
  6: 3, 42, 171,  228,  105,   18,    1;
  7: 3, 51, 279,  579,  465,  153,   21,   1;
  8: 3, 60, 414, 1200, 1500,  828,  210,  24,  1;
  9: 3, 69, 576, 2172, 3858, 3258, 1344, 276, 27, 1;
%F Formula
The row polynomials satisfy (n-1)*P(n,x) = (n-1+(3*n^2-7*n+4)*x-(2*n-2)*x^2)*P(n-1,x) - ((n-1)*x+(10*n-25)*x^2+(n-4)*x^3)*d/dx P(n-1,x) + (x^2+10*x^3+x^4)*d^2/dx^2 P(n-1,x) for n >= 2, with P(0,x) = 1, P(1,x) = 3 + x.
%t Mathematica
P[0, x_] := 1;
P[1, x_] := 3 + x;
P[n_, x_] := P[n, x] = Expand[Together[((n - 1 + (3*n^2 - 7*n + 4)*x - (2*n - 2)*x^2)*P[n - 1, x] - ((n - 1)*x + (10*n - 25)*x^2 + (n - 4)*x^3)*D[P[n - 1, x], x] + (x^2 + 10*x^3 + x^4)*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; v[2] = 3 + x; for(n = 2, N, my(i = n + 1); v[i] = ((n - 1 + (3*n^2 - 7*n + 4)*x - (2*n - 2)*x^2)*v[i - 1] - ((n - 1)*x + (10*n - 25)*x^2 + (n - 4)*x^3)*deriv(v[i - 1], x) + (x^2 + 10*x^3 + x^4)*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)

Generated with polytool c48629374597.

A249250: reversed triangle, offset 0, from the row P0
%e Example
Triangle begins:
  0: 1;
  1: 1,  3;
  2: 1,  6,   3;
  3: 1,  9,  15,    3;
  4: 1, 12,  36,   24,    3;
  5: 1, 15,  66,   90,   33,    3;
  6: 1, 18, 105,  228,  171,   42,    3;
  7: 1, 21, 153,  465,  579,  279,   51,   3;
  8: 1, 24, 210,  828, 1500, 1200,  414,  60,  3;
  9: 1, 27, 276, 1344, 3258, 3858, 2172, 576, 69, 3;
%F Formula
Row n lists the coefficients of P(n,x) in descending powers of x, where (n-1)*P(n,x) = (n-1+(3*n^2-7*n+4)*x-(2*n-2)*x^2)*P(n-1,x) - ((n-1)*x+(10*n-25)*x^2+(n-4)*x^3)*d/dx P(n-1,x) + (x^2+10*x^3+x^4)*d^2/dx^2 P(n-1,x) for n >= 2, with P(0,x) = 1, P(1,x) = 3 + x.
%t Mathematica
P[0, x_] := 1;
P[1, x_] := 3 + x;
P[n_, x_] := P[n, x] = Expand[Together[((n - 1 + (3*n^2 - 7*n + 4)*x - (2*n - 2)*x^2)*P[n - 1, x] - ((n - 1)*x + (10*n - 25)*x^2 + (n - 4)*x^3)*D[P[n - 1, x], x] + (x^2 + 10*x^3 + x^4)*D[P[n - 1, x], {x, 2}])/(n - 1)]];
Table[Reverse[PadRight[CoefficientList[P[n, x], x], n + 1]], {n, 0, 9}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N + 1)); v[1] = 1; v[2] = 3 + x; for(n = 2, N, my(i = n + 1); v[i] = ((n - 1 + (3*n^2 - 7*n + 4)*x - (2*n - 2)*x^2)*v[i - 1] - ((n - 1)*x + (10*n - 25)*x^2 + (n - 4)*x^3)*deriv(v[i - 1], x) + (x^2 + 10*x^3 + x^4)*deriv(deriv(v[i - 1], x), x))/(n - 1)); v};
row(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n + 1 - #c)); c = Vecrev(c); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};
T(9)

Generated with polytool c48629374597.

Proved properties

  • Real-rooted. Every row P_n, n >= 0, is zero or real-rooted with all roots in (-inf, 0]
    Every polynomial of the sequence has only real roots, inside the stated interval.
    by the cited steps: e-4b8c77d59939295e
  • Coefficients of one sign. Every nonzero row, n >= 0: the coefficients of P_n have one sign
    All coefficients of each polynomial have the same sign (after x -> -x when alternating).
    derived from the cited steps by a stated theorem: e-4b8c77d59939295e
  • Hadamard kernel (Malo). Rows P_n, n >= 0, are Hadamard kernels (preserve real-rootedness)
    Multiplying the coefficients of any real-rooted polynomial termwise by those of a row (Hadamard product) gives a real-rooted polynomial again.
    derived from the cited steps by a stated theorem: e-4b8c77d59939295e

Proof chains

1 incoming proof step. Each step is expanded back to the roots of the proof DAG, the theorem checks of a node's own rule. JSON

Same fingerprint (1)

Equal up to content, sign, reversal and a small index shift.

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