A257623 → D+1

nonnegative coefficients

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

Recurrence

\(\left(-12 + 10\,x + 5\,n\right)\,P_{n}(x) = \left(168 - 849\,x + 1808\,x^{2} - 1585\,x^{3} + 350\,x^{4} - 120\,n + 645\,n x - 1810\,n x^{2} + 1550\,n x^{3} - 250\,n x^{4} - 150\,n^{2} x + 775\,n^{2} x^{2} - 375\,n^{2} x^{3} - 125\,n^{3} x^{2}\right)\,P_{n-2}(x) + \left(105\,x - 675\,x^{2} + 1345\,x^{3} - 1025\,x^{4} + 250\,x^{5} - 75\,n x + 550\,n x^{2} - 850\,n x^{3} + 375\,n x^{4} - 125\,n^{2} x^{2} + 125\,n^{2} x^{3}\right)\,P_{n-2}'(x) + \left(-117 + 338\,x - 275\,x^{2} + 50\,x^{3} + 55\,n - 215\,n x + 125\,n x^{2} + 50\,n^{2} x\right)\,P_{n-1}(x) + \left(-60\,x + 110\,x^{2} - 50\,x^{3} + 25\,n x - 25\,n x^{2}\right)\,P_{n-1}'(x)\)

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

\(P_{1}(x) = 6 + 3\,x\)

Rows ?

n	x⁰	x¹	x²	x³	x⁴	x⁵	x⁶	x⁷	P_n(1)
0	1								1
1	6	3							9
2	57	66	9						132
3	528	1503	582	27					2640
4	4575	30546	26508	4818	81				66528
5	38058	538275	1000920	401490	39030	243			2018016
6	310053	8554794	31147575	25499220	5669355	313698	729		71495424
7	2500836	127011903	833172462	1284859725	568255080	77248521	2513958	2187	2895564672

All 15 rows

Technical details

Node id: n-96bb9a675a290cdc · revision 0 · JSON
Definition source: derived:one_plus_derivative
Fingerprint: f6-8e3b879cee2d5239

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:         6,           3;
  2:        57,          66,            9;
  3:       528,        1503,          582,            27;
  4:      4575,       30546,        26508,          4818,            81;
  5:     38058,      538275,      1000920,        401490,         39030,           243;
  6:    310053,     8554794,     31147575,      25499220,       5669355,        313698,          729;
  7:   2500836,   127011903,    833172462,    1284859725,     568255080,      77248521,      2513958,        2187;
  8:  20078859,  1803655434,  19997833266,   53651518446,   43664112300,   11719429974,   1032297654,    20124786,      6561;
  9: 160880190, 24872935923, 444552153504, 1944862866564, 2701196241948, 1314591605250, 230557635888, 13642921764, 161037654, 19683;

%F Formula

The row polynomials satisfy (5*n-12+10*x)*P(n,x) = -(120*n-168+(150*n^2-645*n+849)*x+(125*n^3-775*n^2+1810*n-1808)*x^2+(375*n^2-1550*n+1585)*x^3+(250*n-350)*x^4)*P(n-2,x) - ((75*n-105)*x+(125*n^2-550*n+675)*x^2-(125*n^2-850*n+1345)*x^3-(375*n-1025)*x^4-250*x^5)*d/dx P(n-2,x) + (55*n-117+(50*n^2-215*n+338)*x+(125*n-275)*x^2+50*x^3)*P(n-1,x) + ((25*n-60)*x-(25*n-110)*x^2-50*x^3)*d/dx P(n-1,x) for n >= 2, with P(0,x) = 1, P(1,x) = 6 + 3*x.

%t Mathematica

P[0, x_] := 1;
P[1, x_] := 6 + 3*x;
P[n_, x_] := P[n, x] = Expand[Together[(-(120*n - 168 + (150*n^2 - 645*n + 849)*x + (125*n^3 - 775*n^2 + 1810*n - 1808)*x^2 + (375*n^2 - 1550*n + 1585)*x^3 + (250*n - 350)*x^4)*P[n - 2, x] - ((75*n - 105)*x + (125*n^2 - 550*n + 675)*x^2 - (125*n^2 - 850*n + 1345)*x^3 - (375*n - 1025)*x^4 - 250*x^5)*D[P[n - 2, x], x] + (55*n - 117 + (50*n^2 - 215*n + 338)*x + (125*n - 275)*x^2 + 50*x^3)*P[n - 1, x] + ((25*n - 60)*x - (25*n - 110)*x^2 - 50*x^3)*D[P[n - 1, x], x])/(5*n - 12 + 10*x)]];
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] = 6 + 3*x; for(n = 2, N, my(i = n + 1); v[i] = (-(120*n - 168 + (150*n^2 - 645*n + 849)*x + (125*n^3 - 775*n^2 + 1810*n - 1808)*x^2 + (375*n^2 - 1550*n + 1585)*x^3 + (250*n - 350)*x^4)*v[i - 2] - ((75*n - 105)*x + (125*n^2 - 550*n + 675)*x^2 - (125*n^2 - 850*n + 1345)*x^3 - (375*n - 1025)*x^4 - 250*x^5)*deriv(v[i - 2], x) + (55*n - 117 + (50*n^2 - 215*n + 338)*x + (125*n - 275)*x^2 + 50*x^3)*v[i - 1] + ((25*n - 60)*x - (25*n - 110)*x^2 - 50*x^3)*deriv(v[i - 1], x))/(5*n - 12 + 10*x)); 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 Laguerre’s theorem on differential operators
Coefficients of one sign? Every nonzero row, n ≥ 0: the coefficients of P_n have one sign
derived from Laguerre’s theorem on differential operators
Hadamard kernel (Malo)? Rows P_n, n ≥ 0, are Hadamard kernels (preserve real-rootedness)
derived from Laguerre’s theorem on differential operators
Interlacing of consecutive rows? Rows P_(m-1), P_m interlace (weak, unordered) for every m ≥ 1
by Laguerre’s theorem on differential operators

Proof chains ? JSON

Laguerre’s theorem on differential operators [SP14] polytool ? step ›
- from Triangle read by rows: T(n,k) = t(n-k, k), where t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1) and f(n) = 5*n + 3.?
  - Liu–Wang theorem [LW07, Thm 2.1, Cor 2.4] ? step › direct check
  - Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
    
    from Triangle read by rows: T(n,k) = t(n-k, k), where t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1) and f(n) = 5*n + 3.?
    
    ↑ Liu–Wang theorem (above)

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
[SP14] J Schur and G Polya. Über zwei Arten von Faktorenfolgen in der Theorie der algebraischen Gleichungen. J. Reine Angew. Math., 1914.