A023531 → (1/2)D²+2D+1 → x ↦ x+1

nonnegative coefficients

Proved for all n: all roots real, in (−∞, −1] · coefficients of one sign · Hadamard kernel what this means

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

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1314
266113
310189138
4154036121104
5217510060151272
62812622520090181688
7361964415253501262111696

Generated b-files: A084851 (row sums), A124847 (reversed triangle) ?

Computed in your browser ?

Technical details
Node id
n-5a583a8eb9880032 · revision 1 · JSON
Definition source
derived:shift_x_plus_one
Fingerprint
f6-5dbc84ad685f9f2d

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

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:  3,   1;
  2:  6,   6,    1;
  3: 10,  18,    9,    1;
  4: 15,  40,   36,   12,    1;
  5: 21,  75,  100,   60,   15,    1;
  6: 28, 126,  225,  200,   90,   18,   1;
  7: 36, 196,  441,  525,  350,  126,  21,   1;
  8: 45, 288,  784, 1176, 1050,  560, 168,  24,  1;
  9: 55, 405, 1296, 2352, 2646, 1890, 840, 216, 27, 1;
%F Formula
The row polynomials satisfy P(n,x) = (1+3*x+3*x^2+x^3)*P(n-3,x) - (3+6*x+3*x^2)*P(n-2,x) + (3+3*x)*P(n-1,x) for n >= 3, with P(0,x) = 1, P(1,x) = 3 + x, P(2,x) = 6 + 6*x + x^2.
%t Mathematica
P[0, x_] := 1;
P[1, x_] := 3 + x;
P[2, x_] := 6 + 6*x + x^2;
P[n_, x_] := P[n, x] = Expand[(1 + 3*x + 3*x^2 + x^3)*P[n - 3, x] - (3 + 6*x + 3*x^2)*P[n - 2, x] + (3 + 3*x)*P[n - 1, 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] = 3 + x; v[3] = 6 + 6*x + x^2; for(n = 3, N, my(i = n + 1); v[i] = (1 + 3*x + 3*x^2 + x^3)*v[i - 3] - (3 + 6*x + 3*x^2)*v[i - 2] + (3 + 3*x)*v[i - 1]); 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.

A124847: reversed triangle, offset 1, from the row P0
%e Example
Triangle begins:
   1: 1;
   2: 1,  3;
   3: 1,  6,   6;
   4: 1,  9,  18,  10;
   5: 1, 12,  36,  40,   15;
   6: 1, 15,  60, 100,   75,   21;
   7: 1, 18,  90, 200,  225,  126,   28;
   8: 1, 21, 126, 350,  525,  441,  196,   36;
   9: 1, 24, 168, 560, 1050, 1176,  784,  288,  45;
  10: 1, 27, 216, 840, 1890, 2646, 2352, 1296, 405, 55;
%F Formula
Row n lists the coefficients of P(n,x) in descending powers of x, where P(n,x) = (1+3*x+3*x^2+x^3)*P(n-3,x) - (3+6*x+3*x^2)*P(n-2,x) + (3+3*x)*P(n-1,x) for n >= 4, with P(1,x) = 1, P(2,x) = 3 + x, P(3,x) = 6 + 6*x + x^2.
%t Mathematica
P[1, x_] := 1;
P[2, x_] := 3 + x;
P[3, x_] := 6 + 6*x + x^2;
P[n_, x_] := P[n, x] = Expand[(1 + 3*x + 3*x^2 + x^3)*P[n - 3, x] - (3 + 6*x + 3*x^2)*P[n - 2, x] + (3 + 3*x)*P[n - 1, x]];
Table[Reverse[PadRight[CoefficientList[P[n, x], x], n]], {n, 1, 10}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N)); v[1] = 1; v[2] = 3 + x; v[3] = 6 + 6*x + x^2; for(n = 4, N, my(i = n); v[i] = (1 + 3*x + 3*x^2 + x^3)*v[i - 3] - (3 + 6*x + 3*x^2)*v[i - 2] + (3 + 3*x)*v[i - 1]); v};
row(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n - #c)); c = Vecrev(c); c};
T(N) = {my(v = P(N)); concat(vector(#v, i, row(i, v[i])))};
T(10)

Generated with polytool c48629374597.

Proved properties

  • Real-rooted. Every row P_n, n >= 0, is zero or real-rooted with all roots in (-inf, -1]
    Every polynomial of the sequence has only real roots, inside the stated interval.
    by the cited steps: e-ee5cc8292e809076
  • 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-ee5cc8292e809076
  • 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-ee5cc8292e809076

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

Same fingerprint (1)

Equal up to content, sign, reversal and a small index shift.

References

  1. [ASW52] Michael Aissen, I. J. Schoenberg and A. M. Whitney. On the generating functions of totally positive sequences I. Journal d’Analyse Mathématique, 2(1):93–103, 1952. doi:10.1007/BF02786970
  2. [AW24] Christos A. Athanasiadis and David G. Wagner. Veronese sections and interlacing matrices of polynomials and formal power series. arXiv:2404.12989. arXiv
  3. [Edr53] Albert Edrei. Proof of a Conjecture of Schoenberg on the Generating Function of a Totally Positive Sequence. Canadian Journal of Mathematics, 5:86–94, 1953. doi:10.4153/CJM-1953-010-3
  4. [SP14] J Schur and G Polya. Über zwei Arten von Faktorenfolgen in der Theorie der algebraischen Gleichungen. J. Reine Angew. Math., 1914.