A104029 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(-1 - 5\,x - 10\,x^{2} - 9\,x^{3} - 3\,x^{4}\right)\,P_{n-2}(x) + \left(2 + 8\,x + 10\,x^{2} + 4\,x^{3}\right)\,P_{n-1}(x)\)

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

\(P_{1}(x) = 2 + 3\,x\)

Rows ?

n	x⁰	x¹	x²	x³	x⁴	x⁵	x⁶	x⁷	P_n(1)
0	1								1
1	2	3							5
2	3	11	9						23
3	4	25	47	27					103
4	5	46	143	182	81				457
5	6	75	336	693	668	243			2021
6	7	113	676	1981	3052	2369	729		8927
7	8	161	1225	4740	10273	12620	8201	2187	39415

All 15 rows

Technical details

Node id: n-f2c01aaaeddbf0a7 · revision 0 · JSON
Definition source: derived:mobius_x_over_x_plus_one
Fingerprint: f6-8090db1b43d4bbd1

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:  2,   3;
  2:  3,  11,    9;
  3:  4,  25,   47,    27;
  4:  5,  46,  143,   182,    81;
  5:  6,  75,  336,   693,   668,    243;
  6:  7, 113,  676,  1981,  3052,   2369,    729;
  7:  8, 161, 1225,  4740, 10273,  12620,   8201,   2187;
  8:  9, 220, 2058, 10032, 28483,  48864,  49890,  27884,  6561;
  9: 10, 291, 3264, 19398, 68926, 154001, 218356, 190678, 93494, 19683;

%F Formula

The row polynomials satisfy (1+2*x+x^2)*P(n,x) = -(1+5*x+10*x^2+9*x^3+3*x^4)*P(n-2,x) + (2+8*x+10*x^2+4*x^3)*P(n-1,x) for n >= 2, with P(0,x) = 1, P(1,x) = 2 + 3*x.

%t Mathematica

P[0, x_] := 1;
P[1, x_] := 2 + 3*x;
P[n_, x_] := P[n, x] = Expand[Together[(-(1 + 5*x + 10*x^2 + 9*x^3 + 3*x^4)*P[n - 2, x] + (2 + 8*x + 10*x^2 + 4*x^3)*P[n - 1, 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; v[2] = 2 + 3*x; for(n = 2, N, my(i = n + 1); v[i] = (-(1 + 5*x + 10*x^2 + 9*x^3 + 3*x^4)*v[i - 2] + (2 + 8*x + 10*x^2 + 4*x^3)*v[i - 1])/(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 Triangle, read by rows, of pairwise sums of trinomial coefficients (A027907).?
  - Liu–Wang theorem for three-term recurrences [LW07, Thm 2.1, Thm 2.3, Cor 2.4, Prop 3.3] polytool ? step › direct check
  - Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from Triangle read by rows: T(n,k) is the number of ternary strings of length n with k indispensable digits, with 0 <= k <= …?
    
    Liu–Wang theorem for three-term recurrences [LW07, Thm 2.1, Thm 2.3, Cor 2.4, Prop 3.3] polytool ? step › direct check
    
    Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from Triangle, read by rows, of pairwise sums of trinomial coefficients (A027907).?
    
    ↑ Liu–Wang theorem for three-term recurrences (above)

Related ?

Same fingerprint ?

A330381 → 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