A085841 → reverse

nonnegative coefficients

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

Recurrence

\(\left(-1 + 2\,n\right)\,P_{n}(x) = \left(-4 - 3\,x + 8\,n + 6\,n x\right)\,P_{n-1}(x) + \left(16\,x - 4\,x^{2}\right)\,P_{n-1}'(x)\)

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

Rows ?

n	x⁰	x¹	x²	x³	x⁴	x⁵	x⁶	x⁷	P_n(1)
0	1								1
1	4	3							7
2	16	40	5						61
3	64	336	140	7					547
4	256	2304	2016	336	9				4921
5	1024	14080	21120	7392	660	11			44287
6	4096	79872	183040	109824	20592	1144	13		398581
7	16384	430080	1397760	1281280	411840	48048	1820	15	3587227

All 15 rows

Technical details

Node id: n-9130684ebb51c694 · revision 0 · JSON
Definition source: derived:reversal
Fingerprint: f6-2c0f1fe74efc2e1a

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:      4,        3;
  2:     16,       40,        5;
  3:     64,      336,      140,         7;
  4:    256,     2304,     2016,       336,        9;
  5:   1024,    14080,    21120,      7392,      660,       11;
  6:   4096,    79872,   183040,    109824,    20592,     1144,      13;
  7:  16384,   430080,  1397760,   1281280,   411840,    48048,    1820,     15;
  8:  65536,  2228224,  9748480,  12673024,  6223360,  1244672,   99008,   2720,   17;
  9: 262144, 11206656, 63504384, 111132672, 77395968, 23648768, 3224832, 186048, 3876, 19;

%F Formula

The row polynomials satisfy (2*n-1)*P(n,x) = (8*n-4+(6*n-3)*x)*P(n-1,x) + (16*x-4*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[((8*n - 4 + (6*n - 3)*x)*P[n - 1, x] + (16*x - 4*x^2)*D[P[n - 1, x], x])/(2*n - 1)]];
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] = ((8*n - 4 + (6*n - 3)*x)*v[i - 1] + (16*x - 4*x^2)*deriv(v[i - 1], x))/(2*n - 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)

polytool c48629374597

Properties

Real-rooted? Every row P_n, n ≥ 0, is zero or real-rooted with all roots in (−∞, 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 Triangle: row #n has n+1 terms. T(n,m) = 4^m (2n+1)! / ( (2n-2m)! (2m+1)! ).?
  - Liu–Wang theorem for three-term recurrences [LW07, Thm 2.1, Thm 2.3, Cor 2.4, Prop 3.3] polytool ? step › direct check
  - Three-term recurrence criterion [Sze39, §3.3; Fav35] polytool ? step › direct check

Related ?

Same fingerprint ?

A085841: Triangle: row #n has n+1 terms. T(n,m) = 4^m (2n+1)! / ( (2n-2m)! (2m+1)! ).

References

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