Real-rooted atlas

basis-map coefficients of A026729 → basis A007318

palindromic γ-positive nonnegative coefficients only rational roots

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

Recurrence
\(P_{n}(x) = \left(1 + x\right)\,P_{n-1}(x)\)
\(P_{1}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7x8x9x10x11x12x13x14Pn(1)
111
2112
31214
413318
51464116
61510105132
7161520156164
8172135352171128
918285670562881256
10193684126126843691512
1111045120210252210120451011024
1211155165330462462330165551112048
1311266220495792924792495220661214096
14113782867151287171617161287715286781318192
151149136410012002300334323003200210013649114116384

Generated b-files: A000079 (row sums), A007318, A007318 (reversed triangle) ?

Computed in your browser ?

Technical details
Node id
n-a6f555b0e6271a68 · revision 1 · JSON
Definition source
derived:shift_x_plus_one
Fingerprint
f6-77a7ed3b1d5f257e

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 -1)
  • 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): yes

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 = 1, …, ascending powers of x
%e Example
Triangle begins:
   1: 1;
   2: 1, 1;
   3: 1, 2,  1;
   4: 1, 3,  3,  1;
   5: 1, 4,  6,  4,   1;
   6: 1, 5, 10, 10,   5,   1;
   7: 1, 6, 15, 20,  15,   6,  1;
   8: 1, 7, 21, 35,  35,  21,  7,  1;
   9: 1, 8, 28, 56,  70,  56, 28,  8, 1;
  10: 1, 9, 36, 84, 126, 126, 84, 36, 9, 1;
%F Formula
The row polynomials satisfy P(n,x) = (1+x)*P(n-1,x) for n >= 2, with P(1,x) = 1.
%t Mathematica
P[1, x_] := 1;
P[n_, x_] := P[n, x] = Expand[(1 + x)*P[n - 1, x]];
Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N)); v[1] = 1; for(n = 2, N, my(i = n); v[i] = (1 + 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, v[i])))};
T(10)

Generated with polytool c48629374597.

A007318: triangle, offset 0, from the row P1
%e Example
Triangle begins:
  0: 1;
  1: 1, 1;
  2: 1, 2,  1;
  3: 1, 3,  3,  1;
  4: 1, 4,  6,  4,   1;
  5: 1, 5, 10, 10,   5,   1;
  6: 1, 6, 15, 20,  15,   6,  1;
  7: 1, 7, 21, 35,  35,  21,  7,  1;
  8: 1, 8, 28, 56,  70,  56, 28,  8, 1;
  9: 1, 9, 36, 84, 126, 126, 84, 36, 9, 1;
%F Formula
The row polynomials satisfy P(n,x) = (1+x)*P(n-1,x) for n >= 1, with P(0,x) = 1.
%t Mathematica
P[0, x_] := 1;
P[n_, x_] := P[n, x] = Expand[(1 + x)*P[n - 1, 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; for(n = 1, N, my(i = n + 1); v[i] = (1 + x)*v[i - 1]); 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.

A007318: reversed triangle, offset 0, from the row P1
%e Example
Triangle begins:
  0: 1;
  1: 1, 1;
  2: 1, 2,  1;
  3: 1, 3,  3,  1;
  4: 1, 4,  6,  4,   1;
  5: 1, 5, 10, 10,   5,   1;
  6: 1, 6, 15, 20,  15,   6,  1;
  7: 1, 7, 21, 35,  35,  21,  7,  1;
  8: 1, 8, 28, 56,  70,  56, 28,  8, 1;
  9: 1, 9, 36, 84, 126, 126, 84, 36, 9, 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) for n >= 1, with P(0,x) = 1.
%t Mathematica
P[0, x_] := 1;
P[n_, x_] := P[n, x] = Expand[(1 + x)*P[n - 1, 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; for(n = 1, N, my(i = n + 1); v[i] = (1 + x)*v[i - 1]); 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 >= 1, 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-aefe61e44b642324
  • Coefficients of one sign. Every nonzero row, n >= 1: 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-279353e75dd63f59, e-aefe61e44b642324
  • Hadamard kernel (Malo). Rows P_n, n >= 1, 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-279353e75dd63f59, e-aefe61e44b642324
  • Real-rooted. Every row P_n, n >= 1, 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-279353e75dd63f59

Proof chains

2 incoming proof steps. Each step is expanded back to the roots of the proof DAG, the theorem checks of a node's own rule. JSON

Same fingerprint (59)

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