A185410 → reverse → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

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

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

Rows ?

n	x⁰	x¹	x²	x³	x⁴	x⁵	x⁶	x⁷	P_n(1)
0	1								1
1	0	1							1
2	0	2	3						5
3	0	4	18	15					37
4	0	8	84	180	105				377
5	0	16	360	1500	2100	945			4921
6	0	32	1488	10800	27300	28350	10395		78365
7	0	64	6048	72240	294000	529200	436590	135135	1473277

All 15 rows

Technical details

Node id: n-dba95db7d17b96cd · revision 0 · JSON
Definition source: derived:mobius_x_over_x_plus_one
Fingerprint: f6-f2b1ecb4ea267ced

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: 0,   1;
  2: 0,   2,     3;
  3: 0,   4,    18,      15;
  4: 0,   8,    84,     180,      105;
  5: 0,  16,   360,    1500,     2100,       945;
  6: 0,  32,  1488,   10800,    27300,     28350,     10395;
  7: 0,  64,  6048,   72240,   294000,    529200,    436590,    135135;
  8: 0, 128, 24384,  463680,  2857680,   7938000,  11060280,   7567560,   2027025;
  9: 0, 256, 97920, 2904000, 26107200, 105099120, 220041360, 249729480, 145945800, 34459425;

%F Formula

The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (x+2*x^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 >= 1, with P(0,x) = 1.

%t Mathematica

P[0, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((x + 2*x^2 + x^3)*P[n - 1, x] + (2*x + 6*x^2 + 6*x^3 + 2*x^4)*D[P[n - 1, x], x])/(1 + 2*x + x^2)]];
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] = ((x + 2*x^2 + x^3)*v[i - 1] + (2*x + 6*x^2 + 6*x^3 + 2*x^4)*deriv(v[i - 1], x))/(1 + 2*x + x^2)); 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 A triangular decomposition of the double factorial numbers A001147.?
  - Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from A decomposition of the double factorials A001147.?
    
    Liu–Wang theorem [LW07, Thm 2.1, Cor 2.4] ? step › direct check
    
    Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from A triangular decomposition of the double factorial numbers A001147.?
    
    Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from A decomposition of the double factorials A001147.?
    
    ↑ Liu–Wang theorem (above)
  - ↑ Affine, reversal and Möbius substitutions (elementary) (above)

Related ?

Same fingerprint ?

A185410 → x ↦ x+1

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