A154228: Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)*(n+2)*(2*n+3)/6)*T(n-2, k-1), read by rows.
A154228 palindromic γ-positive nonnegative coefficients
Proved for all n: all roots real, in (−∞, 0) · coefficients of one sign · interlacing · Hadamard kernel what this means
Rows ?
| n | x0 | x1 | x2 | x3 | x4 | x5 | x6 | x7 | Pn(1) |
|---|---|---|---|---|---|---|---|---|---|
| 0 | 1 | 1 | |||||||
| 1 | 1 | 1 | 2 | ||||||
| 2 | 1 | 16 | 1 | 18 | |||||
| 3 | 1 | 47 | 47 | 1 | 96 | ||||
| 4 | 1 | 103 | 974 | 103 | 1 | 1182 | |||
| 5 | 1 | 195 | 5354 | 5354 | 195 | 1 | 11100 | ||
| 6 | 1 | 336 | 19969 | 147068 | 19969 | 336 | 1 | 187680 | |
| 7 | 1 | 541 | 60085 | 1259253 | 1259253 | 60085 | 541 | 1 | 2639760 |
Generated b-files: A154228, A154228 (reversed triangle) ?
Computed in your browser ?
Technical details
- Node id
n-a3274460eb3b9452· revision 1 · JSON- Definition source
oeis_catalog:A154228- Fingerprint
f6-318b826a84c258d5
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
OEIS matches ?
- A154228: reversed triangle, 55 terms agree, 10 full rows row-aligned (tabl) OEIS nonn in polytool catalog · view at the OEIS · generated b-file 990 terms from n = 0 ?
- A154228: triangle, 55 terms agree, 10 full rows row-aligned (tabl) OEIS nonn in polytool catalog · view at the OEIS · generated b-file 990 terms from n = 0 ?
A generated b-file is computed from the node's definition along the row alignment of its OEIS link and starts at the entry's offset. It is an identification on a finite prefix, not a proof that the node and the OEIS entry agree for every n.
OEIS submission text ?
Fields to paste into an OEIS submission: the example %e, the formula
%F and programs %t (Mathematica) and %o (PARI/GP) that list the same rows.
For an OEIS link the row numbers, the offset and the orientation are those of the entry, as matched;
the matched terms were compared with the published ones, which identifies the node with the entry on a
finite prefix, not for every n. The PARI/GP program is not machine-checked here.
The atlas's own rows: n = 0, …, ascending powers of x
%e ExampleTriangle begins: 0: 1; 1: 1, 1; 2: 1, 16, 1; 3: 1, 47, 47, 1; 4: 1, 103, 974, 103, 1; 5: 1, 195, 5354, 5354, 195, 1; 6: 1, 336, 19969, 147068, 19969, 336, 1; 7: 1, 541, 60085, 1259253, 1259253, 60085, 541, 1; 8: 1, 827, 156386, 7010503, 44432886, 7010503, 156386, 827, 1; 9: 1, 1213, 365498, 30299614, 536255794, 536255794, 30299614, 365498, 1213, 1;
%F FormulaThe row polynomials satisfy P(n,x) = (1+x)*P(n-1,x) + ((1/3)*n^3+(3/2)*n^2+(13/6)*n+1)*x*P(n-2,x) for n >= 2, with P(0,x) = 1, P(1,x) = 1 + x.
%t MathematicaP[0, x_] := 1;
P[1, x_] := 1 + x;
P[n_, x_] := P[n, x] = Expand[(1 + x)*P[n - 1, x] + ((1/3)*n^3 + (3/2)*n^2 + (13/6)*n + 1)*x*P[n - 2, 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 + x; for(n = 2, N, my(i = n + 1); v[i] = (1 + x)*v[i - 1] + ((1/3)*n^3 + (3/2)*n^2 + (13/6)*n + 1)*x*v[i - 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)Generated with polytool c48629374597.
A154228: triangle, offset 0, from the row P0
%e ExampleTriangle begins: 0: 1; 1: 1, 1; 2: 1, 16, 1; 3: 1, 47, 47, 1; 4: 1, 103, 974, 103, 1; 5: 1, 195, 5354, 5354, 195, 1; 6: 1, 336, 19969, 147068, 19969, 336, 1; 7: 1, 541, 60085, 1259253, 1259253, 60085, 541, 1; 8: 1, 827, 156386, 7010503, 44432886, 7010503, 156386, 827, 1; 9: 1, 1213, 365498, 30299614, 536255794, 536255794, 30299614, 365498, 1213, 1;
%F FormulaThe row polynomials satisfy P(n,x) = (1+x)*P(n-1,x) + ((1/3)*n^3+(3/2)*n^2+(13/6)*n+1)*x*P(n-2,x) for n >= 2, with P(0,x) = 1, P(1,x) = 1 + x.
%t MathematicaP[0, x_] := 1;
P[1, x_] := 1 + x;
P[n_, x_] := P[n, x] = Expand[(1 + x)*P[n - 1, x] + ((1/3)*n^3 + (3/2)*n^2 + (13/6)*n + 1)*x*P[n - 2, x]];
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; for(n = 2, N, my(i = n + 1); v[i] = (1 + x)*v[i - 1] + ((1/3)*n^3 + (3/2)*n^2 + (13/6)*n + 1)*x*v[i - 2]); 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)Generated with polytool c48629374597.
A154228: reversed triangle, offset 0, from the row P0
%e ExampleTriangle begins: 0: 1; 1: 1, 1; 2: 1, 16, 1; 3: 1, 47, 47, 1; 4: 1, 103, 974, 103, 1; 5: 1, 195, 5354, 5354, 195, 1; 6: 1, 336, 19969, 147068, 19969, 336, 1; 7: 1, 541, 60085, 1259253, 1259253, 60085, 541, 1; 8: 1, 827, 156386, 7010503, 44432886, 7010503, 156386, 827, 1; 9: 1, 1213, 365498, 30299614, 536255794, 536255794, 30299614, 365498, 1213, 1;
%F FormulaRow n lists the coefficients of P(n,x) in descending powers of x, where P(n,x) = (1+x)*P(n-1,x) + ((1/3)*n^3+(3/2)*n^2+(13/6)*n+1)*x*P(n-2,x) for n >= 2, with P(0,x) = 1, P(1,x) = 1 + x.
%t MathematicaP[0, x_] := 1;
P[1, x_] := 1 + x;
P[n_, x_] := P[n, x] = Expand[(1 + x)*P[n - 1, x] + ((1/3)*n^3 + (3/2)*n^2 + (13/6)*n + 1)*x*P[n - 2, x]];
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; for(n = 2, N, my(i = n + 1); v[i] = (1 + x)*v[i - 1] + ((1/3)*n^3 + (3/2)*n^2 + (13/6)*n + 1)*x*v[i - 2]); 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)Generated with polytool c48629374597.
Proved properties
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Real-rooted. Every row P_n, n >= 0, is zero or real-rooted with all roots in (-inf, 0)
Every polynomial of the sequence has only real roots, inside the stated interval.by the cited steps:
e-a3c76982c9b6ad3f -
Coefficients of one sign. Every nonzero row, n >= 0: the coefficients of P_n have one sign
All coefficients of each polynomial have the same sign (after x -> -x when alternating).derived from the cited steps by a stated theorem:
e-a3c76982c9b6ad3f -
Hadamard kernel (Malo). Rows P_n, n >= 0, are Hadamard kernels (preserve real-rootedness)
Multiplying the coefficients of any real-rooted polynomial termwise by those of a row (Hadamard product) gives a real-rooted polynomial again.derived from the cited steps by a stated theorem:
e-a3c76982c9b6ad3f -
Interlacing of consecutive rows. Rows P_(m-1), P_m interlace (strict, previous interlaces next) for every m >= 1
The roots of consecutive polynomials alternate on the real line.by the cited steps:
e-a3c76982c9b6ad3f
Proof chains
1 incoming proof step. Each step is expanded back to the roots of the proof DAG, the theorem checks of a node's own rule. JSON
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theorem_check liu_wang_derivative_lag
e-a3c76982c9b6ad3froot: theorem checkLiu–Wang theorem with a lag row [LW07, Thm 2.1] RealRooted ?
Used in 6 proof steps
e-1820edefae4adbcfbasis_map (A013611) → basis_map_A013611 of Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)*(n+2)*(2*n+3)/6)…e-239c141e75570afcbasis_map (A008277) → basis_map_A008277 of Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)*(n+2)*(2*n+3)/6)…e-2967c46d8636e83fbasis_map (A007318) → basis_map_A007318 of Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)*(n+2)*(2*n+3)/6)…e-48c6fe64c9c46a35basis_map (A013609) → basis_map_A013609 of Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)*(n+2)*(2*n+3)/6)…e-bbc3c7bdcf5b3d8abasis_map (A013610) → basis_map_A013610 of Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)*(n+2)*(2*n+3)/6)…e-92fa46b1f7f7427egamma_polynomial (scale 1, shift 0) → gamma_polynomial of Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)*(n+2)*(2*n+3)/6)*…
Related sequences ?
No candidate equalities or related sequences.
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