A102662 → basis A013609

nonnegative coefficients

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

Recurrence
\(P_{n}(x) = \left(1 + 2\,x\right)\,P_{n-2}(x) + \left(2 + 2\,x\right)\,P_{n-1}(x)\)
\(P_{0}(x) = 1\)
\(P_{1}(x) = 4 + 6\,x\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
14610
29221243
322768024202
45323636823248937
512869814401384608964354
63091994511666164496150419220227
7746556017056277282560013312358438493970

Computed in your browser ?

Technical details
Node id
n-3dddab772a6867d2 · revision 1 · JSON
Definition source
derived:basis_map_A013609
Fingerprint
f6-5104a571b27debd0

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 ?

No OEIS matches.

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:    4,     6;
  2:    9,    22,     12;
  3:   22,    76,     80,     24;
  4:   53,   236,    368,    232,     48;
  5:  128,   698,   1440,   1384,    608,     96;
  6:  309,  1994,   5116,   6616,   4496,   1504,    192;
  7:  746,  5560,  17056,  27728,  25600,  13312,   3584,   384;
  8: 1801, 15224,  54336, 106416, 124384,  88320,  36992,  8320,   768;
  9: 4348, 41102, 167296, 383344, 542656, 489920, 280832, 98176, 18944, 1536;
%F Formula
The row polynomials satisfy P(n,x) = (1+2*x)*P(n-2,x) + (2+2*x)*P(n-1,x) for n >= 2, with P(0,x) = 1, P(1,x) = 4 + 6*x.
%t Mathematica
P[0, x_] := 1;
P[1, x_] := 4 + 6*x;
P[n_, x_] := P[n, x] = Expand[(1 + 2*x)*P[n - 2, x] + (2 + 2*x)*P[n - 1, x]];
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] = 4 + 6*x; for(n = 2, N, my(i = n + 1); v[i] = (1 + 2*x)*v[i - 2] + (2 + 2*x)*v[i - 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.

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-30bb9f6b8a15c298
  • 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-30bb9f6b8a15c298
  • 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-30bb9f6b8a15c298
  • Interlacing of consecutive rows. Rows P_(m-1), P_m interlace (weak, unordered) for every m >= 1
    The roots of consecutive polynomials alternate on the real line.
    by the cited steps: e-30bb9f6b8a15c298

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

No candidate equalities or related sequences.

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