A144442 → x ↦ x+1
nonnegative coefficients
Proved for all n all roots real, in (−∞, −1]coefficients of one signHadamard kernelinterlacing (weak)
Recurrence
\(P_{n}(x) = \left(5 + 5\,x\right)\,P_{n-2}(x) + \left(-3 - 4\,x + 5\,n + 5\,n x\right)\,P_{n-1}(x) + \left(-5\,x - 5\,x^{2}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)
\(P_{1}(x) = 2 + x\)
Rows ?
| n | x0 | x1 | x2 | x3 | x4 | x5 | x6 | x7 | Pn(1) |
|---|---|---|---|---|---|---|---|---|---|
| 0 | 1 | 1 | |||||||
| 1 | 2 | 1 | 3 | ||||||
| 2 | 19 | 19 | 1 | 39 | |||||
| 3 | 238 | 357 | 121 | 1 | 717 | ||||
| 4 | 4141 | 8282 | 4874 | 733 | 1 | 18031 | |||
| 5 | 92292 | 230730 | 193390 | 59355 | 4405 | 1 | 580173 | ||
| 6 | 2512589 | 7537767 | 8198740 | 3834535 | 687410 | 26437 | 1 | 22797479 | |
| 7 | 80864308 | 283025078 | 378474822 | 238624360 | 69920280 | 7768599 | 158629 | 1 | 1058836077 |
Compute more
?
Technical details
- Node id
n-21a70536fc2023cb· revision 0 · JSON- Definition source
derived:shift_x_plus_one- Fingerprint
f6-83f6dcff4e22986d
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 ExampleTriangle begins: 0: 1; 1: 2, 1; 2: 19, 19, 1; 3: 238, 357, 121, 1; 4: 4141, 8282, 4874, 733, 1; 5: 92292, 230730, 193390, 59355, 4405, 1; 6: 2512589, 7537767, 8198740, 3834535, 687410, 26437, 1; 7: 80864308, 283025078, 378474822, 238624360, 69920280, 7768599, 158629, 1; 8: 3004542341, 12018169364, 19071280147, 15150247667, 6222366045, 1215516903, 86697182, 951781, 1; 9: 126595099862, 569677949379, 1046242561308, 1003351867476, 532341215532, 151721918691, 20528272772, 961124619, 5710693, 1;
%F FormulaThe row polynomials satisfy P(n,x) = (5+5*x)*P(n-2,x) + (5*n-3+(5*n-4)*x)*P(n-1,x) - (5*x+5*x^2)*d/dx P(n-1,x) for n >= 2, with P(0,x) = 1, P(1,x) = 2 + x.
%t MathematicaP[0, x_] := 1;
P[1, x_] := 2 + x;
P[n_, x_] := P[n, x] = Expand[(5 + 5*x)*P[n - 2, x] + (5*n - 3 + (5*n - 4)*x)*P[n - 1, x] - (5*x + 5*x^2)*D[P[n - 1, x], x]];
Table[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten%o PARI/GPP(N) = {my(v = vector(N + 1)); v[1] = 1; v[2] = 2 + x; for(n = 2, N, my(i = n + 1); v[i] = (5 + 5*x)*v[i - 2] + (5*n - 3 + (5*n - 4)*x)*v[i - 1] - (5*x + 5*x^2)*deriv(v[i - 1], 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 (−∞, -1]
- Coefficients of one sign? Every nonzero row, n ≥ 0: the coefficients of P_n have one sign
- Hadamard kernel (Malo)? Rows P_n, n ≥ 0, are Hadamard kernels (preserve real-rootedness)
- Interlacing of consecutive rows? Rows P_(m-1), P_m interlace (weak, unordered) for every m ≥ 1
Proof chains ? JSON
-
Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
-
from
Triangle read by rows: T(n, k) = (5*n-5*k+1)*T(n-1, k-1) +(5*k-4)*T(n-1, k) + 5*T(n-2, k-1) with T(n, 1) = T(n, n) = 1.?
-
Liu–Wang theorem with a lag row [LW07, Thm 2.1] RealRooted ? step › direct check
-
-
from
Triangle read by rows: T(n, k) = (5*n-5*k+1)*T(n-1, k-1) +(5*k-4)*T(n-1, k) + 5*T(n-2, k-1) with T(n, 1) = T(n, n) = 1.?
Related ?
Same fingerprint ?
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