Real-rooted atlas

basis-map coefficients of A026729 → basis A056242 → reverse

nonnegative coefficients

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

Recurrence
\(\left(-1 + n\right)\,P_{n}(x) = \left(-2 - 5\,x + 2\,n + 3\,n x\right)\,P_{n-1}(x) + \left(-2\,x - 2\,x^{2}\right)\,P_{n-1}'(x)\)
\(P_{1}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
111
2213
345110
48169134
5164441141116
63211214685201396
7642724563771552711352
8128640131214088332593514616

Show all 15 stored rows

Generated b-files: A007052 (row sums), A048580 (row sums), A056242 (reversed triangle), A117317 ?

Computed in your browser ?

Technical details
Node id
n-f268708db386b729 · revision 1 · JSON
Definition source
derived:reversal
Fingerprint
f6-74459ae06326a5fd

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 -1)
  • 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 = 1, …, ascending powers of x
%e Example
Triangle begins:
   1:   1;
   2:   2,    1;
   3:   4,    5,    1;
   4:   8,   16,    9,     1;
   5:  16,   44,   41,    14,     1;
   6:  32,  112,  146,    85,    20,    1;
   7:  64,  272,  456,   377,   155,   27,    1;
   8: 128,  640, 1312,  1408,   833,  259,   35,   1;
   9: 256, 1472, 3568,  4712,  3649, 1652,  406,  44,  1;
  10: 512, 3328, 9312, 14608, 14002, 8361, 3024, 606, 54, 1;
%F Formula
The row polynomials satisfy (n-1)*P(n,x) = (2*n-2+(3*n-5)*x)*P(n-1,x) - (2*x+2*x^2)*d/dx P(n-1,x) for n >= 2, with P(1,x) = 1.
%t Mathematica
P[1, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((2*n - 2 + (3*n - 5)*x)*P[n - 1, x] - (2*x + 2*x^2)*D[P[n - 1, x], x])/(n - 1)]];
Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N)); v[1] = 1; for(n = 2, N, my(i = n); v[i] = ((2*n - 2 + (3*n - 5)*x)*v[i - 1] - (2*x + 2*x^2)*deriv(v[i - 1], x))/(n - 1)); 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)

Generated with polytool c48629374597.

A056242: reversed triangle, offset 1, from the row P1
%e Example
Triangle begins:
   1: 1;
   2: 1,  2;
   3: 1,  5,   4;
   4: 1,  9,  16,    8;
   5: 1, 14,  41,   44,   16;
   6: 1, 20,  85,  146,  112,    32;
   7: 1, 27, 155,  377,  456,   272,    64;
   8: 1, 35, 259,  833, 1408,  1312,   640,  128;
   9: 1, 44, 406, 1652, 3649,  4712,  3568, 1472,  256;
  10: 1, 54, 606, 3024, 8361, 14002, 14608, 9312, 3328, 512;
%F Formula
Row n lists the coefficients of P(n,x) in descending powers of x, where (n-1)*P(n,x) = (2*n-2+(3*n-5)*x)*P(n-1,x) - (2*x+2*x^2)*d/dx P(n-1,x) for n >= 2, with P(1,x) = 1.
%t Mathematica
P[1, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((2*n - 2 + (3*n - 5)*x)*P[n - 1, x] - (2*x + 2*x^2)*D[P[n - 1, x], x])/(n - 1)]];
Table[Reverse[PadRight[CoefficientList[P[n, x], x], n]], {n, 1, 10}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N)); v[1] = 1; for(n = 2, N, my(i = n); v[i] = ((2*n - 2 + (3*n - 5)*x)*v[i - 1] - (2*x + 2*x^2)*deriv(v[i - 1], x))/(n - 1)); v};
row(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n - #c)); c = Vecrev(c); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i, v[i])))};
T(10)

Generated with polytool c48629374597.

A117317: triangle, offset 0, from the row P1
%e Example
Triangle begins:
  0:   1;
  1:   2,    1;
  2:   4,    5,    1;
  3:   8,   16,    9,     1;
  4:  16,   44,   41,    14,     1;
  5:  32,  112,  146,    85,    20,    1;
  6:  64,  272,  456,   377,   155,   27,    1;
  7: 128,  640, 1312,  1408,   833,  259,   35,   1;
  8: 256, 1472, 3568,  4712,  3649, 1652,  406,  44,  1;
  9: 512, 3328, 9312, 14608, 14002, 8361, 3024, 606, 54, 1;
%F Formula
The row polynomials satisfy n*P(n,x) = (2*n+(3*n-2)*x)*P(n-1,x) - (2*x+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[((2*n + (3*n - 2)*x)*P[n - 1, x] - (2*x + 2*x^2)*D[P[n - 1, x], x])/(n)]];
Table[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] = ((2*n + (3*n - 2)*x)*v[i - 1] - (2*x + 2*x^2)*deriv(v[i - 1], x))/(n)); v};
row(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n + 1 - #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 >= 1, 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-3cf80f21f053e148
  • Coefficients of one sign. Every nonzero row, n >= 1: 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-3cf80f21f053e148
  • Hadamard kernel (Malo). Rows P_n, n >= 1, 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-3cf80f21f053e148

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.