Real-rooted atlas

A075500 → reverse

nonnegative coefficients

Proved for all n all roots real, in (−∞, 0]coefficients of one signHadamard kernel

Recurrence

\(P_{n}(x) = \left(1 + 10\,x + 5\,n x\right)\,P_{n-1}(x) + \left(-5\,x^{2}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1 + 15\,x + 25\,x^{2}\)

Rows ?

nx0x1x2x3x4x5x6x7x8x9Pn(1)
01152541
1130175125331
215062518756253176
317516251125019375312535451
4110535004375018812519687515625447981
511406650131250106312530187501984375781256282416
6118011550330750434437524281250472656251992187539062596546231
71225187507350001426687513289062553289062572890625019960937519531251611270851

All 15 rows

Technical details
Node id
n-7f851af1bfd10b81 · revision 0 · JSON
Definition source
derived:reversal
Fingerprint
f6-9e37acaa9ec1379f

Observed on the stored rows ?

  • Degrees: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 (affine: slope 1, intercept 2)
  • 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

Matches ?

No OEIS matches.

Submission text ?

Atlas rows n = 0, …, ascending powers of x
%e Example
Triangle begins:
  0: 1,  15,    25;
  1: 1,  30,   175,     125;
  2: 1,  50,   625,    1875,      625;
  3: 1,  75,  1625,   11250,    19375,       3125;
  4: 1, 105,  3500,   43750,   188125,     196875,       15625;
  5: 1, 140,  6650,  131250,  1063125,    3018750,     1984375,        78125;
  6: 1, 180, 11550,  330750,  4344375,   24281250,    47265625,     19921875,       390625;
  7: 1, 225, 18750,  735000, 14266875,  132890625,   532890625,    728906250,    199609375,      1953125;
  8: 1, 275, 28875, 1485000, 39991875,  560896875,  3855156250,  11386718750,  11133203125,   1998046875,     9765625;
  9: 1, 330, 42625, 2784375, 99391875, 1960612500, 20682062500, 107765625000, 238867578125, 168996093750, 19990234375, 48828125;
%F Formula
The row polynomials satisfy P(n,x) = (1+(5*n+10)*x)*P(n-1,x) - 5*x^2*d/dx P(n-1,x) for n >= 1, with P(0,x) = 1 + 15*x + 25*x^2.
%t Mathematica
P[0, x_] := 1 + 15*x + 25*x^2;
P[n_, x_] := P[n, x] = Expand[(1 + (5*n + 10)*x)*P[n - 1, x] - 5*x^2*D[P[n - 1, x], x]];
Table[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N + 1)); v[1] = 1 + 15*x + 25*x^2; for(n = 1, N, my(i = n + 1); v[i] = (1 + (5*n + 10)*x)*v[i - 1] - 5*x^2*deriv(v[i - 1], x)); 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

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.