A096940: Pascal (1,5) triangle.

A096940 ↗ nonnegative coefficientsonly rational roots

Proved for all n all roots real, in (−∞, 0)coefficients of one signHadamard kernelinterlacing (weak)

Recurrence from A096940

\(\left(-1 + n\right)\,P_{n}(x) = \left(-1 - x + n + n x\right)\,P_{n-1}(x)\)
\(P_{0}(x) = 5\)
\(P_{1}(x) = 1 + 5\,x\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
055
1156
216512
31711524
4181816548
519263421596
6110356055265192
7111459511581315384
Technical details
Node id
n-6fc8b9c5d9c18144 · revision 1 · JSON
Definition source
oeis_catalog:A096940
Fingerprint
f6-5bcf47c8067a6574

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): yes

Matches ?

  • A096940 triangle 72 terms, 11 rows · tabl · nonn · catalog b-file?

Row sums ? A003945, A167881, A111286, A042950, A058764, A007283, A091629, A082505

Submission text ?

Atlas rows n = 0, …, ascending powers of x
%e Example
Triangle begins:
  0: 5;
  1: 1,  5;
  2: 1,  6,  5;
  3: 1,  7, 11,   5;
  4: 1,  8, 18,  16,   5;
  5: 1,  9, 26,  34,  21,   5;
  6: 1, 10, 35,  60,  55,  26,   5;
  7: 1, 11, 45,  95, 115,  81,  31,   5;
  8: 1, 12, 56, 140, 210, 196, 112,  36,  5;
  9: 1, 13, 68, 196, 350, 406, 308, 148, 41, 5;
%F Formula
The row polynomials satisfy (n-1)*P(n,x) = (n-1+(n-1)*x)*P(n-1,x) for n >= 2, with P(0,x) = 5, P(1,x) = 1 + 5*x.
%t Mathematica
P[0, x_] := 5;
P[1, x_] := 1 + 5*x;
P[n_, x_] := P[n, x] = Expand[Together[((n - 1 + (n - 1)*x)*P[n - 1, x])/(n - 1)]];
Table[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N + 1)); v[1] = 5; v[2] = 1 + 5*x; for(n = 2, N, my(i = n + 1); v[i] = ((n - 1 + (n - 1)*x)*v[i - 1])/(n - 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)

polytool c48629374597

A096940 triangle, offset 0, from P0
%e Example
Triangle begins:
  0: 5;
  1: 1,  5;
  2: 1,  6,  5;
  3: 1,  7, 11,   5;
  4: 1,  8, 18,  16,   5;
  5: 1,  9, 26,  34,  21,   5;
  6: 1, 10, 35,  60,  55,  26,   5;
  7: 1, 11, 45,  95, 115,  81,  31,   5;
  8: 1, 12, 56, 140, 210, 196, 112,  36,  5;
  9: 1, 13, 68, 196, 350, 406, 308, 148, 41, 5;
%F Formula
The row polynomials satisfy (n-1)*P(n,x) = (n-1+(n-1)*x)*P(n-1,x) for n >= 2, with P(0,x) = 5, P(1,x) = 1 + 5*x.
%t Mathematica
P[0, x_] := 5;
P[1, x_] := 1 + 5*x;
P[n_, x_] := P[n, x] = Expand[Together[((n - 1 + (n - 1)*x)*P[n - 1, x])/(n - 1)]];
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] = 5; v[2] = 1 + 5*x; for(n = 2, N, my(i = n + 1); v[i] = ((n - 1 + (n - 1)*x)*v[i - 1])/(n - 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)

polytool c48629374597

Properties

Proof chains ? JSON

Same fingerprint ?

References

  1. [Bra06] Petter Brändén. On linear transformations preserving the Pólya frequency property. Transactions of the American Mathematical Society, 358(8):3697–3716, February 2006. doi:10.1090/s0002-9947-06-03856-6
  2. [Bra15] Petter Brändén. Unimodality, log-concavity, real-rootedness and beyond. Handbook of enumerative combinatorics, :437–483, March 2015. doi:10.1201/b18255-10
  3. [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
  4. [SP14] J Schur and G Polya. Über zwei Arten von Faktorenfolgen in der Theorie der algebraischen Gleichungen. J. Reine Angew. Math., 1914.