Real-rooted atlas

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

Recurrence from OEIS A154228
\(P_{n}(x) = \left(1 + x\right)\,P_{n-1}(x) + \left(x + \frac{13}{6}\,n x + \frac{3}{2}\,n^{2} x + \frac{1}{3}\,n^{3} x\right)\,P_{n-2}(x)\)
\(P_{0}(x) = 1\)
\(P_{1}(x) = 1 + x\)

Rows ?

nx0x1x2x3x4x5x6x7x8x9x10x11x12x13x14Pn(1)
011
1112
2116118
314747196
4110397410311182
5119553545354195111100
6133619969147068199693361187680
7154160085125925312592536008554112639760
818271563867010503444328867010503156386827158768320
9112133654983029961453625579453625579430299614365498121311133844240
101172078517310979642841138699262355555190441138699261097964287851731720132004458400
111237115753433481553012391841545437623568793037623568793023918415454348155301157534323711801007672800
12131912986394992787331114189845287376941357277820044468385236376941357277811418984528799278733129863943191127813666775200
13142075396150259474687046856026313328160795103875405693105206964405693105206964281607951038754685602631332594746870539615042071868650121442400
141544893571976303271580170221130044317022476352288851079267305555592566652700810656851079267305555591702247635228881702211300443630327158093571975448136226247044132800

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 ?

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 Example
Triangle 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 Formula
The 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 Mathematica
P[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/GP
P(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 Example
Triangle 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 Formula
The 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 Mathematica
P[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/GP
P(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 Example
Triangle 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 Formula
Row 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 Mathematica
P[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/GP
P(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

  • 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

Used in 6 proof steps

No candidate equalities or related sequences.

References

  1. [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