A062196 → reverse

nonnegative coefficients

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

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

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1314
268115
3103015156
4158090241210
521175350210351792
628336105011204204813003
73658826464410294075663111440

Generated b-files: A062196 (reversed triangle) ?

Computed in your browser ?

Technical details
Node id
n-3f65f4dc3d5d123d · revision 1 · JSON
Definition source
derived:reversal
Fingerprint
f6-5f98bd4aea48ed7d

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:  6,    8,     1;
  3: 10,   30,    15,     1;
  4: 15,   80,    90,    24,     1;
  5: 21,  175,   350,   210,    35,     1;
  6: 28,  336,  1050,  1120,   420,    48,     1;
  7: 36,  588,  2646,  4410,  2940,   756,    63,    1;
  8: 45,  960,  5880, 14112, 14700,  6720,  1260,   80,  1;
  9: 55, 1485, 11880, 38808, 58212, 41580, 13860, 1980, 99, 1;
%F Formula
The row polynomials satisfy (n^2+2*n)*P(n,x) = (n^2+4*n+4+(3*n^2+2*n-2)*x)*P(n-1,x) + ((2*n+2)*x-(2*n+2)*x^2)*d/dx 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[((n^2 + 4*n + 4 + (3*n^2 + 2*n - 2)*x)*P[n - 1, x] + ((2*n + 2)*x - (2*n + 2)*x^2)*D[P[n - 1, x], x])/(n^2 + 2*n)]];
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] = ((n^2 + 4*n + 4 + (3*n^2 + 2*n - 2)*x)*v[i - 1] + ((2*n + 2)*x - (2*n + 2)*x^2)*deriv(v[i - 1], x))/(n^2 + 2*n)); 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.

A062196: reversed triangle, offset 0, from the row P0
%e Example
Triangle begins:
  0: 1;
  1: 1,  3;
  2: 1,  8,    6;
  3: 1, 15,   30,    10;
  4: 1, 24,   90,    80,    15;
  5: 1, 35,  210,   350,   175,    21;
  6: 1, 48,  420,  1120,  1050,   336,    28;
  7: 1, 63,  756,  2940,  4410,  2646,   588,    36;
  8: 1, 80, 1260,  6720, 14700, 14112,  5880,   960,   45;
  9: 1, 99, 1980, 13860, 41580, 58212, 38808, 11880, 1485, 55;
%F Formula
Row n lists the coefficients of P(n,x) in descending powers of x, where (n^2+2*n)*P(n,x) = (n^2+4*n+4+(3*n^2+2*n-2)*x)*P(n-1,x) + ((2*n+2)*x-(2*n+2)*x^2)*d/dx 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[((n^2 + 4*n + 4 + (3*n^2 + 2*n - 2)*x)*P[n - 1, x] + ((2*n + 2)*x - (2*n + 2)*x^2)*D[P[n - 1, x], x])/(n^2 + 2*n)]];
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; for(n = 1, N, my(i = n + 1); v[i] = ((n^2 + 4*n + 4 + (3*n^2 + 2*n - 2)*x)*v[i - 1] + ((2*n + 2)*x - (2*n + 2)*x^2)*deriv(v[i - 1], x))/(n^2 + 2*n)); 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-b6d3c37eae668953
  • 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-b6d3c37eae668953
  • 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-b6d3c37eae668953

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