A116414 → basis A007318 → reverse

nonnegative coefficients

Proved for all n all roots real, in (−1, 0]coefficients of one sign Hadamard kernel

Recurrence

\(\left(2\,n\right)\,P_{n}(x) = \left(-5\,x - 13\,x^{2} + 2\,n + 15\,n x + 13\,n x^{2}\right)\,P_{n-1}(x) + \left(-x - 10\,x^{2} - 13\,x^{3}\right)\,P_{n-1}'(x)\)

\(P_{0}(x) = 1\)

Rows ?

n	x⁰	x¹	x²	x³	x⁴	x⁵	x⁶	x⁷	P_n(1)
0	1								1
1	1	5							6
2	1	10	22						33
3	1	15	69	95					180
4	1	20	141	410	409				981
5	1	25	238	1070	2252	1760			5346
6	1	30	360	2200	7179	11790	7573		29133
7	1	35	507	3925	17465	44475	59767	32585	158760

All 15 rows

Technical details

Node id: n-4c39ee03cea0e59f · revision 0 · JSON
Definition source: derived:reversal
Fingerprint: f6-1cfe01b94310c21a

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

Matches ?

No OEIS matches.

Submission text ?

Atlas rows n = 0, …, ascending powers of x

%e Example

Triangle begins:
  0: 1;
  1: 1,  5;
  2: 1, 10,  22;
  3: 1, 15,  69,   95;
  4: 1, 20, 141,  410,   409;
  5: 1, 25, 238, 1070,  2252,   1760;
  6: 1, 30, 360, 2200,  7179,  11790,   7573;
  7: 1, 35, 507, 3925, 17465,  44475,  59767,   32585;
  8: 1, 40, 679, 6370, 36010, 125200, 260605,  296050,  140206;
  9: 1, 45, 876, 9660, 66339, 293475, 834210, 1465650, 1441155, 603275;

%F Formula

The row polynomials satisfy 2*n*P(n,x) = (2*n+(15*n-5)*x+(13*n-13)*x^2)*P(n-1,x) - (x+10*x^2+13*x^3)*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 + (15*n - 5)*x + (13*n - 13)*x^2)*P[n - 1, x] - (x + 10*x^2 + 13*x^3)*D[P[n - 1, x], x])/(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] = ((2*n + (15*n - 5)*x + (13*n - 13)*x^2)*v[i - 1] - (x + 10*x^2 + 13*x^3)*deriv(v[i - 1], x))/(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)

polytool c48629374597

Properties

Real-rooted? Every row P_n, n ≥ 0, is zero or real-rooted with all roots in (-1, 0]
by Affine, reversal and Möbius substitutions (elementary)
Coefficients of one sign? Every nonzero row, n ≥ 0: the coefficients of P_n have one sign
derived from Affine, reversal and Möbius substitutions (elementary)
Hadamard kernel (Malo)? Rows P_n, n ≥ 0, are Hadamard kernels (preserve real-rootedness)
derived from Affine, reversal and Möbius substitutions (elementary)

Proof chains ? JSON

Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
- from basis_map_A007318 of Riordan array (1/((1-x)(1-3x)),x/((1-x)(1-3x))).?
  - Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from Riordan array (1/((1-x)(1-3x)),x/((1-x)(1-3x))).?
    
    Three-term recurrence criterion [Sze39, §3.3; Fav35] polytool ? step › direct check
  - Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from Riordan array (1/((1-x)(1-3x)),x/((1-x)(1-3x))).?
    
    ↑ Three-term recurrence criterion (above)

Related ?

Candidate equalities ?

A116414 → reverse → x ↦ x/(x+1) 15 rows; no all-index equality certificate was found

Same fingerprint ?

A116414 → basis A007318

References

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