A146988: Triangle, read by rows, T(n, k) = binomial(n, k) for n < 2 and binomial(n, k) + 4^(n-1) * binomial(n-2, k-1) otherwise.
A146988 ↗ palindromicγ-positivenonnegative coefficients
Proved for all n all roots real, in (−∞, 0)coefficients of one signHadamard kernelinterlacing (weak)
Recurrence from A146988
\(\left(-3 + n\right)\,P_{n}(x) = \left(-15 - 15\,x + 5\,n + 5\,n x\right)\,P_{n-1}(x) + \left(12 + 24\,x + 12\,x^{2} - 4\,n - 8\,n x - 4\,n x^{2}\right)\,P_{n-2}(x)\)
\(P_{0}(x) = 1\)
\(P_{1}(x) = 1 + x\)
\(P_{2}(x) = 1 + 6\,x + x^{2}\)
\(P_{3}(x) = 1 + 19\,x + 19\,x^{2} + x^{3}\)
Rows ?
| n | x0 | x1 | x2 | x3 | x4 | x5 | x6 | x7 | Pn(1) |
|---|---|---|---|---|---|---|---|---|---|
| 0 | 1 | 1 | |||||||
| 1 | 1 | 1 | 2 | ||||||
| 2 | 1 | 6 | 1 | 8 | |||||
| 3 | 1 | 19 | 19 | 1 | 40 | ||||
| 4 | 1 | 68 | 134 | 68 | 1 | 272 | |||
| 5 | 1 | 261 | 778 | 778 | 261 | 1 | 2080 | ||
| 6 | 1 | 1030 | 4111 | 6164 | 4111 | 1030 | 1 | 16448 | |
| 7 | 1 | 4103 | 20501 | 40995 | 40995 | 20501 | 4103 | 1 | 131200 |
All 15 rows b-files: A146988, A146988 (reversed triangle) ?
Compute more
?
Technical details
- Node id
n-3f596f3a08c8e69f· revision 0 · JSON- Definition source
oeis_catalog:A146988- Fingerprint
f6-771cff9e4c79260d
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: yes
- γ-positive (gamma_positive): yes
- Nonnegative coefficients (
nonn): yes - Only rational roots (
rr_by_factors: every nonzero row is a product of linear factors over ℚ): no
Matches ?
- A146988 triangle 55 terms, 10 rows · tabl · nonn · catalog b-file?
- A146988 reversed triangle 55 terms, 10 rows · tabl · nonn · catalog b-file?
Submission text ?
Atlas rows n = 0, …, ascending powers of x
%e ExampleTriangle begins: 0: 1; 1: 1, 1; 2: 1, 6, 1; 3: 1, 19, 19, 1; 4: 1, 68, 134, 68, 1; 5: 1, 261, 778, 778, 261, 1; 6: 1, 1030, 4111, 6164, 4111, 1030, 1; 7: 1, 4103, 20501, 40995, 40995, 20501, 4103, 1; 8: 1, 16392, 98332, 245816, 327750, 245816, 98332, 16392, 1; 9: 1, 65545, 458788, 1376340, 2293886, 2293886, 1376340, 458788, 65545, 1;
%F FormulaThe row polynomials satisfy (n-3)*P(n,x) = (5*n-15+(5*n-15)*x)*P(n-1,x) - (4*n-12+(8*n-24)*x+(4*n-12)*x^2)*P(n-2,x) for n >= 4, with P(0,x) = 1, P(1,x) = 1 + x, P(2,x) = 1 + 6*x + x^2, P(3,x) = 1 + 19*x + 19*x^2 + x^3.
%t MathematicaP[0, x_] := 1;
P[1, x_] := 1 + x;
P[2, x_] := 1 + 6*x + x^2;
P[3, x_] := 1 + 19*x + 19*x^2 + x^3;
P[n_, x_] := P[n, x] = Expand[Together[((5*n - 15 + (5*n - 15)*x)*P[n - 1, x] - (4*n - 12 + (8*n - 24)*x + (4*n - 12)*x^2)*P[n - 2, x])/(n - 3)]];
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 + x; v[3] = 1 + 6*x + x^2; v[4] = 1 + 19*x + 19*x^2 + x^3; for(n = 4, N, my(i = n + 1); v[i] = ((5*n - 15 + (5*n - 15)*x)*v[i - 1] - (4*n - 12 + (8*n - 24)*x + (4*n - 12)*x^2)*v[i - 2])/(n - 3)); 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
A146988 triangle, offset 0, from P0
%e ExampleTriangle begins: 0: 1; 1: 1, 1; 2: 1, 6, 1; 3: 1, 19, 19, 1; 4: 1, 68, 134, 68, 1; 5: 1, 261, 778, 778, 261, 1; 6: 1, 1030, 4111, 6164, 4111, 1030, 1; 7: 1, 4103, 20501, 40995, 40995, 20501, 4103, 1; 8: 1, 16392, 98332, 245816, 327750, 245816, 98332, 16392, 1; 9: 1, 65545, 458788, 1376340, 2293886, 2293886, 1376340, 458788, 65545, 1;
%F FormulaThe row polynomials satisfy (n-3)*P(n,x) = (5*n-15+(5*n-15)*x)*P(n-1,x) - (4*n-12+(8*n-24)*x+(4*n-12)*x^2)*P(n-2,x) for n >= 4, with P(0,x) = 1, P(1,x) = 1 + x, P(2,x) = 1 + 6*x + x^2, P(3,x) = 1 + 19*x + 19*x^2 + x^3.
%t MathematicaP[0, x_] := 1;
P[1, x_] := 1 + x;
P[2, x_] := 1 + 6*x + x^2;
P[3, x_] := 1 + 19*x + 19*x^2 + x^3;
P[n_, x_] := P[n, x] = Expand[Together[((5*n - 15 + (5*n - 15)*x)*P[n - 1, x] - (4*n - 12 + (8*n - 24)*x + (4*n - 12)*x^2)*P[n - 2, x])/(n - 3)]];
Table[PadRight[CoefficientList[P[n, x], x], n + 1], {n, 0, 9}] // Flatten%o PARI/GPP(N) = {my(v = vector(N + 1)); v[1] = 1; v[2] = 1 + x; v[3] = 1 + 6*x + x^2; v[4] = 1 + 19*x + 19*x^2 + x^3; for(n = 4, N, my(i = n + 1); v[i] = ((5*n - 15 + (5*n - 15)*x)*v[i - 1] - (4*n - 12 + (8*n - 24)*x + (4*n - 12)*x^2)*v[i - 2])/(n - 3)); v};
row(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n + 1 - #c)); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};
T(9)polytool c48629374597
A146988 reversed triangle, offset 0, from P0
%e ExampleTriangle begins: 0: 1; 1: 1, 1; 2: 1, 6, 1; 3: 1, 19, 19, 1; 4: 1, 68, 134, 68, 1; 5: 1, 261, 778, 778, 261, 1; 6: 1, 1030, 4111, 6164, 4111, 1030, 1; 7: 1, 4103, 20501, 40995, 40995, 20501, 4103, 1; 8: 1, 16392, 98332, 245816, 327750, 245816, 98332, 16392, 1; 9: 1, 65545, 458788, 1376340, 2293886, 2293886, 1376340, 458788, 65545, 1;
%F FormulaRow n lists the coefficients of P(n,x) in descending powers of x, where (n-3)*P(n,x) = (5*n-15+(5*n-15)*x)*P(n-1,x) - (4*n-12+(8*n-24)*x+(4*n-12)*x^2)*P(n-2,x) for n >= 4, with P(0,x) = 1, P(1,x) = 1 + x, P(2,x) = 1 + 6*x + x^2, P(3,x) = 1 + 19*x + 19*x^2 + x^3.
%t MathematicaP[0, x_] := 1;
P[1, x_] := 1 + x;
P[2, x_] := 1 + 6*x + x^2;
P[3, x_] := 1 + 19*x + 19*x^2 + x^3;
P[n_, x_] := P[n, x] = Expand[Together[((5*n - 15 + (5*n - 15)*x)*P[n - 1, x] - (4*n - 12 + (8*n - 24)*x + (4*n - 12)*x^2)*P[n - 2, x])/(n - 3)]];
Table[Reverse[PadRight[CoefficientList[P[n, x], x], n + 1]], {n, 0, 9}] // Flatten%o PARI/GPP(N) = {my(v = vector(N + 1)); v[1] = 1; v[2] = 1 + x; v[3] = 1 + 6*x + x^2; v[4] = 1 + 19*x + 19*x^2 + x^3; for(n = 4, N, my(i = n + 1); v[i] = ((5*n - 15 + (5*n - 15)*x)*v[i - 1] - (4*n - 12 + (8*n - 24)*x + (4*n - 12)*x^2)*v[i - 2])/(n - 3)); v};
row(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n + 1 - #c)); c = Vecrev(c); 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)
-
Coefficients of one sign?
Every nonzero row, n ≥ 0: the coefficients of P_n have one sign
derived from Liu–Wang theorem for three-term recurrences
-
Hadamard kernel (Malo)?
Rows P_n, n ≥ 0, are Hadamard kernels (preserve real-rootedness)
derived from Liu–Wang theorem for three-term recurrences
- Interlacing of consecutive rows? Rows P_(m-1), P_m interlace (weak, previous interlaces next) for every m ≥ 3
Proof chains ? JSON
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