A272866 → basis A007318 → reverse

nonnegative coefficients

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

Recurrence

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

Rows ?

nx0x1x2x3x4x5x6x7x8x9x10x11x12x13x14Pn(1)
011
115511
2110355025121
3115902754503751251331
4120170800227540004250250062514641
51252751750712519375356254375034375156253125161051
61304053250172506375016812531875043125040625025312593750156251771561
71355605425355251662505731251478125286562541562504440625339062517500005468757812519487171
Technical details
Node id
n-9d7ba545bc982df0 · revision 0 · JSON
Definition source
derived:reversal
Fingerprint
f6-d62194962de0e59f

Observed on the stored rows ?

  • Degrees: 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28 (affine: slope 2, 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

Matches ?

No OEIS matches.

Submission text ?

Atlas rows n = 0, …, ascending powers of x
%e Example
Triangle begins:
  0: 1;
  1: 1,  5,   5;
  2: 1, 10,  35,    50,     25;
  3: 1, 15,  90,   275,    450,    375,     125;
  4: 1, 20, 170,   800,   2275,   4000,    4250,     2500,      625;
  5: 1, 25, 275,  1750,   7125,  19375,   35625,    43750,    34375,     15625,      3125;
  6: 1, 30, 405,  3250,  17250,  63750,  168125,   318750,   431250,    406250,    253125,     93750,     15625;
  7: 1, 35, 560,  5425,  35525, 166250,  573125,  1478125,  2865625,   4156250,   4440625,   3390625,   1750000,    546875,     78125;
  8: 1, 40, 740,  8400,  65450, 371000, 1582000,  5175000, 13121875,  25875000,  39550000,  46375000,  40906250,  26250000,  11562500,   3125000,   390625;
  9: 1, 45, 945, 12300, 111150, 740250, 3764250, 14940000, 46906875, 117359375, 234534375, 373500000, 470531250, 462656250, 347343750, 192187500, 73828125, 17578125, 1953125;
%F Formula
The row polynomials satisfy P(n,x) = (1+5*x+5*x^2)*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 + 5*x + 5*x^2)*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; for(n = 1, N, my(i = n + 1); v[i] = (1 + 5*x + 5*x^2)*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)

polytool c48629374597

Properties

Proof chains ? JSON

Candidate equalities ?

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. [SP14] J Schur and G Polya. Über zwei Arten von Faktorenfolgen in der Theorie der algebraischen Gleichungen. J. Reine Angew. Math., 1914.