A063007 → x ↦ x+1 → reverse

nonnegative coefficients

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

Recurrence

\(\left(-1 + n\right)\,P_{n}(x) = \left(-6 - 15\,x - 8\,x^{2} + 4\,n + 9\,n x + 4\,n x^{2}\right)\,P_{n-1}(x) + \left(-2\,x - 6\,x^{2} - 4\,x^{3}\right)\,P_{n-1}'(x)\)

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

Rows ?

n	x⁰	x¹	x²	x³	x⁴	x⁵	x⁶	x⁷	P_n(1)
1	1								1
2	2	3							5
3	6	18	13						37
4	20	90	132	63					305
5	70	420	930	900	321				2641
6	252	1890	5600	8190	5910	1683			23525
7	924	8316	30870	60480	65940	37926	8989		213445
8	3432	36036	160776	395010	577080	501228	239624	48639	1961825

All 15 rows

Technical details

Node id: n-a882d0a2ea16bc46 · revision 0 · JSON
Definition source: derived:reversal
Fingerprint: f6-c277ed2a39f7c92c

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

Matches ?

No OEIS matches.

Submission text ?

Atlas rows n = 1, …, ascending powers of x

%e Example

Triangle begins:
   1:     1;
   2:     2,      3;
   3:     6,     18,      13;
   4:    20,     90,     132,       63;
   5:    70,    420,     930,      900,      321;
   6:   252,   1890,    5600,     8190,     5910,     1683;
   7:   924,   8316,   30870,    60480,    65940,    37926,     8989;
   8:  3432,  36036,  160776,   395010,   577080,   501228,   239624,    48639;
   9: 12870, 154440,  804804,  2378376,  4358970,  5072760,  3660300,  1497096,  265729;
  10: 48620, 656370, 3912480, 13513500, 29801772, 43513470, 42060480, 25951860, 9274410, 1462563;

%F Formula

The row polynomials satisfy (n-1)*P(n,x) = (4*n-6+(9*n-15)*x+(4*n-8)*x^2)*P(n-1,x) - (2*x+6*x^2+4*x^3)*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[((4*n - 6 + (9*n - 15)*x + (4*n - 8)*x^2)*P[n - 1, x] - (2*x + 6*x^2 + 4*x^3)*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] = ((4*n - 6 + (9*n - 15)*x + (4*n - 8)*x^2)*v[i - 1] - (2*x + 6*x^2 + 4*x^3)*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)

polytool c48629374597

Properties

Real-rooted? Every row P_n, n ≥ 1, 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 ≥ 1: the coefficients of P_n have one sign
derived from Affine, reversal and Möbius substitutions (elementary)
Hadamard kernel (Malo)? Rows P_n, n ≥ 1, 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 T(n,k) = binomial(n,k)*binomial(n+k,k), 0 <= k <= n, triangle read by rows.?
  - Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from T(n,k) = binomial(n,k)*binomial(n+k,k), 0 <= k <= n, triangle read by rows.?
    
    Three-term recurrence criterion [Sze39, §3.3; Fav35] polytool ? step › direct check
  - Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from T(n,k) = binomial(n,k)*binomial(n+k,k), 0 <= k <= n, triangle read by rows.?
    
    ↑ Three-term recurrence criterion (above)

Related ?

Candidate equalities ?

A104684 → x ↦ x/(x+1) 15 rows; no all-index equality certificate was found

Same fingerprint ?

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