A199478 → x ↦ x+1

nonnegative coefficients

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

Recurrence

\(\left(16 + 8\,n + n^{2}\right)\,P_{n}(x) = \left(32 + 16\,x + 25\,n + 20\,n x + 11\,n^{2} + 4\,n^{2} x\right)\,P_{n-1}(x) + \left(-32\,x - 32\,x^{2} - 36\,n - 41\,n x - 8\,n x^{2}\right)\,P_{n-1}'(x) + \left(27 + 51\,x + 24\,x^{2}\right)\,P_{n-1}''(x)\)

\(P_{0}(x) = 2 + x\)

Rows ?

n	x⁰	x¹	x²	x³	x⁴	P_n(1)
0	2	1				3
1	4	3				7
2	8	9	2			19
3	16	23	8			47
4	32	57	30	4		123
5	64	135	92	20		311
6	128	313	266	88	8	803
7	256	711	720	312	48	2047

All 15 rows

Technical details

Node id: n-6dacb02b1cf8b2b9 · revision 0 · JSON
Definition source: derived:shift_x_plus_one
Fingerprint: f6-0a329c2f7da3bd08

Observed on the stored rows ?

Degrees: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8 (half affine: slope 1, intercept 2)
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:    2,    1;
  1:    4,    3;
  2:    8,    9,    2;
  3:   16,   23,    8;
  4:   32,   57,   30,    4;
  5:   64,  135,   92,   20;
  6:  128,  313,  266,   88,   8;
  7:  256,  711,  720,  312,  48;
  8:  512, 1593, 1878, 1020, 240,  16;
  9: 1024, 3527, 4740, 3084, 960, 112;

%F Formula

The row polynomials satisfy (n^2+8*n+16)*P(n,x) = (11*n^2+25*n+32+(4*n^2+20*n+16)*x)*P(n-1,x) - (36*n+(41*n+32)*x+(8*n+32)*x^2)*d/dx P(n-1,x) + (27+51*x+24*x^2)*d^2/dx^2 P(n-1,x) for n >= 1, with P(0,x) = 2 + x.

%t Mathematica

P[0, x_] := 2 + x;
P[n_, x_] := P[n, x] = Expand[Together[((11*n^2 + 25*n + 32 + (4*n^2 + 20*n + 16)*x)*P[n - 1, x] - (36*n + (41*n + 32)*x + (8*n + 32)*x^2)*D[P[n - 1, x], x] + (27 + 51*x + 24*x^2)*D[P[n - 1, x], {x, 2}])/(n^2 + 8*n + 16)]];
Table[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten

%o PARI/GP

P(N) = {my(v = vector(N + 1)); v[1] = 2 + x; for(n = 1, N, my(i = n + 1); v[i] = ((11*n^2 + 25*n + 32 + (4*n^2 + 20*n + 16)*x)*v[i - 1] - (36*n + (41*n + 32)*x + (8*n + 32)*x^2)*deriv(v[i - 1], x) + (27 + 51*x + 24*x^2)*deriv(deriv(v[i - 1], x), x))/(n^2 + 8*n + 16)); 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]
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 Irregular triangle read by rows: T(n,i) = 2^(i-1)*(binomial(n-i+1,i)+binomial(n-i,i)), n >= 1, 0 <= i <= (n+1)/2.?
  - Liu–Wang theorem for three-term recurrences [LW07, Thm 2.1, Thm 2.3, Cor 2.4, Prop 3.3] polytool ? step › direct check

Related ?

Candidate equalities ?

A199478 → basis A007318 15 rows; no all-index equality certificate was found

Same fingerprint ?

A209599 → basis A013609

References

[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