A154233 → x ↦ x/(x+1)
nonnegative coefficients
Proved for all n all roots real, in [−1, 0]coefficients of one signHadamard kernel
Recurrence
\(\left(1 + x\right)\,P_{n}(x) = \left(-x - 2\,x^{2} - x^{3} + n x + 2\,n x^{2} + n x^{3} + n^{2} x + 2\,n^{2} x^{2} + n^{2} x^{3}\right)\,P_{n-2}(x) + \left(1 + 3\,x + 2\,x^{2}\right)\,P_{n-1}(x)\)
\(P_{0}(x) = 1\)
\(P_{1}(x) = 1 + 2\,x\)
Rows ?
| n | x0 | x1 | x2 | x3 | x4 | x5 | x6 | x7 | Pn(1) |
|---|---|---|---|---|---|---|---|---|---|
| 0 | 1 | 1 | |||||||
| 1 | 1 | 2 | 3 | ||||||
| 2 | 1 | 9 | 9 | 19 | |||||
| 3 | 1 | 22 | 60 | 40 | 123 | ||||
| 4 | 1 | 43 | 294 | 502 | 251 | 1091 | |||
| 5 | 1 | 74 | 1047 | 3468 | 4155 | 1662 | 10407 | ||
| 6 | 1 | 117 | 2999 | 19379 | 43727 | 40845 | 13615 | 120683 | |
| 7 | 1 | 174 | 7358 | 87032 | 330810 | 547564 | 415240 | 118640 | 1506819 |
Compute more
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Technical details
- Node id
n-3c6f163a8ebc0747· revision 0 · JSON- Definition source
derived:mobius_x_over_x_plus_one- Fingerprint
f6-ec0cbe08708f3ce3
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: 1, 2; 2: 1, 9, 9; 3: 1, 22, 60, 40; 4: 1, 43, 294, 502, 251; 5: 1, 74, 1047, 3468, 4155, 1662; 6: 1, 117, 2999, 19379, 43727, 40845, 13615; 7: 1, 174, 7358, 87032, 330810, 547564, 415240, 118640; 8: 1, 247, 16084, 322984, 2093712, 5689710, 7514980, 4815780, 1203945; 9: 1, 338, 32153, 1025500, 11140390, 47065072, 97069686, 105535296, 58350825, 12966850;
%F FormulaThe row polynomials satisfy (1+x)*P(n,x) = ((n^2+n-1)*x+(2*n^2+2*n-2)*x^2+(n^2+n-1)*x^3)*P(n-2,x) + (1+3*x+2*x^2)*P(n-1,x) for n >= 2, with P(0,x) = 1, P(1,x) = 1 + 2*x.
%t MathematicaP[0, x_] := 1;
P[1, x_] := 1 + 2*x;
P[n_, x_] := P[n, x] = Expand[Together[(((n^2 + n - 1)*x + (2*n^2 + 2*n - 2)*x^2 + (n^2 + n - 1)*x^3)*P[n - 2, x] + (1 + 3*x + 2*x^2)*P[n - 1, x])/(1 + 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] = 1 + 2*x; for(n = 2, N, my(i = n + 1); v[i] = (((n^2 + n - 1)*x + (2*n^2 + 2*n - 2)*x^2 + (n^2 + n - 1)*x^3)*v[i - 2] + (1 + 3*x + 2*x^2)*v[i - 1])/(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, 0]
- 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)
Proof chains ? JSON
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Affine, reversal and Möbius substitutions (elementary) polytool ? step ›
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from
Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + (n^2 +n -1)*T(n-2, k-1), read by rows.?
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Liu–Wang theorem with a lag row [LW07, Thm 2.1] RealRooted ? step › direct check
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from
Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + (n^2 +n -1)*T(n-2, k-1), read by rows.?
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