Real-rooted atlas

narayana → x ↦ x/(x+1)

nonnegative coefficients

Proved for all n: all roots real, in [−1, 0] · coefficients of one sign · Hadamard kernel what this means

Recurrence
\(\left(1 + 2\,x + x^{2} + n + 2\,n x + n x^{2}\right)\,P_{n}(x) = \left(1 + 4\,x + 5\,x^{2} + 2\,x^{3} + n + 4\,n x + 5\,n x^{2} + 2\,n x^{3}\right)\,P_{n-1}(x) + \left(2\,x + 6\,x^{2} + 6\,x^{3} + 2\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{1}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
111
2123
315511
419211445
5114568442197
6120120300330132903
7127225825148512874294279
81353851925500570075005143020793

Show all 15 stored rows

Generated b-files: A033282, A126216 (reversed triangle) ?

Computed in your browser ?

Technical details
Node id
n-4d61144188f12208 · revision 1 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-1e5fbd8fe1b88630

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: 1,  2;
   3: 1,  5,   5;
   4: 1,  9,  21,   14;
   5: 1, 14,  56,   84,    42;
   6: 1, 20, 120,  300,   330,   132;
   7: 1, 27, 225,  825,  1485,  1287,    429;
   8: 1, 35, 385, 1925,  5005,  7007,   5005,   1430;
   9: 1, 44, 616, 4004, 14014, 28028,  32032,  19448,  4862;
  10: 1, 54, 936, 7644, 34398, 91728, 148512, 143208, 75582, 16796;
%F Formula
The row polynomials satisfy (n+1+(2*n+2)*x+(n+1)*x^2)*P(n,x) = (n+1+(4*n+4)*x+(5*n+5)*x^2+(2*n+2)*x^3)*P(n-1,x) + (2*x+6*x^2+6*x^3+2*x^4)*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[((n + 1 + (4*n + 4)*x + (5*n + 5)*x^2 + (2*n + 2)*x^3)*P[n - 1, x] + (2*x + 6*x^2 + 6*x^3 + 2*x^4)*D[P[n - 1, x], x])/(n + 1 + (2*n + 2)*x + (n + 1)*x^2)]];
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] = ((n + 1 + (4*n + 4)*x + (5*n + 5)*x^2 + (2*n + 2)*x^3)*v[i - 1] + (2*x + 6*x^2 + 6*x^3 + 2*x^4)*deriv(v[i - 1], x))/(n + 1 + (2*n + 2)*x + (n + 1)*x^2)); 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.

A033282: triangle, offset 3, from the row P1
%e Example
Triangle begins:
   3: 1;
   4: 1,  2;
   5: 1,  5,   5;
   6: 1,  9,  21,   14;
   7: 1, 14,  56,   84,    42;
   8: 1, 20, 120,  300,   330,   132;
   9: 1, 27, 225,  825,  1485,  1287,    429;
  10: 1, 35, 385, 1925,  5005,  7007,   5005,   1430;
  11: 1, 44, 616, 4004, 14014, 28028,  32032,  19448,  4862;
  12: 1, 54, 936, 7644, 34398, 91728, 148512, 143208, 75582, 16796;
%F Formula
The row polynomials satisfy (n-1+(2*n-2)*x+(n-1)*x^2)*P(n,x) = (n-1+(4*n-4)*x+(5*n-5)*x^2+(2*n-2)*x^3)*P(n-1,x) + (2*x+6*x^2+6*x^3+2*x^4)*d/dx P(n-1,x) for n >= 4, with P(3,x) = 1.
%t Mathematica
P[3, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((n - 1 + (4*n - 4)*x + (5*n - 5)*x^2 + (2*n - 2)*x^3)*P[n - 1, x] + (2*x + 6*x^2 + 6*x^3 + 2*x^4)*D[P[n - 1, x], x])/(n - 1 + (2*n - 2)*x + (n - 1)*x^2)]];
Table[PadRight[CoefficientList[P[n, x], x], n - 2], {n, 3, 12}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N - 2)); v[1] = 1; for(n = 4, N, my(i = n - 2); v[i] = ((n - 1 + (4*n - 4)*x + (5*n - 5)*x^2 + (2*n - 2)*x^3)*v[i - 1] + (2*x + 6*x^2 + 6*x^3 + 2*x^4)*deriv(v[i - 1], x))/(n - 1 + (2*n - 2)*x + (n - 1)*x^2)); v};
row(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n - 2 - #c)); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i + 2, v[i])))};
T(12)

Generated with polytool c48629374597.

A126216: reversed triangle, offset 1, from the row P1
%e Example
Triangle begins:
   1:     1;
   2:     2,     1;
   3:     5,     5,      1;
   4:    14,    21,      9,      1;
   5:    42,    84,     56,     14,     1;
   6:   132,   330,    300,    120,    20,     1;
   7:   429,  1287,   1485,    825,   225,    27,    1;
   8:  1430,  5005,   7007,   5005,  1925,   385,   35,   1;
   9:  4862, 19448,  32032,  28028, 14014,  4004,  616,  44,  1;
  10: 16796, 75582, 143208, 148512, 91728, 34398, 7644, 936, 54, 1;
%F Formula
Row n lists the coefficients of P(n,x) in descending powers of x, where (n+1+(2*n+2)*x+(n+1)*x^2)*P(n,x) = (n+1+(4*n+4)*x+(5*n+5)*x^2+(2*n+2)*x^3)*P(n-1,x) + (2*x+6*x^2+6*x^3+2*x^4)*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[((n + 1 + (4*n + 4)*x + (5*n + 5)*x^2 + (2*n + 2)*x^3)*P[n - 1, x] + (2*x + 6*x^2 + 6*x^3 + 2*x^4)*D[P[n - 1, x], x])/(n + 1 + (2*n + 2)*x + (n + 1)*x^2)]];
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] = ((n + 1 + (4*n + 4)*x + (5*n + 5)*x^2 + (2*n + 2)*x^3)*v[i - 1] + (2*x + 6*x^2 + 6*x^3 + 2*x^4)*deriv(v[i - 1], x))/(n + 1 + (2*n + 2)*x + (n + 1)*x^2)); 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.

Proved properties

  • Real-rooted. Every row P_n, n >= 1, is zero or real-rooted with all roots in [-1, 0]
    Every polynomial of the sequence has only real roots, inside the stated interval.
    by the cited steps: e-50c973bfb8d0ccf1
  • 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-50c973bfb8d0ccf1
  • 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-50c973bfb8d0ccf1

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 (6)

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

References

  1. [Fav35] Jean Favard. Sur les polynômes de tchebicheff. C. R. Acad. Sci. Paris, 200:2052–2053, 1935.
  2. [Sze39] G. Szegő. Orthogonal Polynomials. Colloquium publications, American Mathematical Society, 1939. doi:10.1090/coll/023