Real-rooted atlas

basis-map coefficients of A112227 → basis A013611 → x ↦ x/(x+1)

nonnegative coefficientsonly rational roots

Proved for all n all roots real, in [−1, 0]coefficients of one signHadamard kernel

Recurrence

\(\left(1 + x\right)\,P_{n}(x) = \left(1 + 6\,x + 5\,x^{2}\right)\,P_{n-1}(x)\)
\(P_{1}(x) = 1 + 10\,x + 25\,x^{2}\)

Rows ?

nx0x1x2x3x4x5x6x7x8x9Pn(1)
11102536
211575125216
31201505006251296
41252501250312531257776
513037525009375187501562546656
61355254375218756562510937578125279936
71407007000437501750004375006250003906251679616
81459001050078750393750131250028125003515625195312510077696

All 15 rows

Technical details
Node id
n-af03db3dc7849a56 · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-a8e18dd2f0709b33

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 1)
  • 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 ?

No OEIS matches.

Submission text ?

Atlas rows n = 1, …, ascending powers of x
%e Example
Triangle begins:
   1: 1, 10,   25;
   2: 1, 15,   75,   125;
   3: 1, 20,  150,   500,    625;
   4: 1, 25,  250,  1250,   3125,    3125;
   5: 1, 30,  375,  2500,   9375,   18750,   15625;
   6: 1, 35,  525,  4375,  21875,   65625,  109375,    78125;
   7: 1, 40,  700,  7000,  43750,  175000,  437500,   625000,   390625;
   8: 1, 45,  900, 10500,  78750,  393750, 1312500,  2812500,  3515625,   1953125;
   9: 1, 50, 1125, 15000, 131250,  787500, 3281250,  9375000, 17578125,  19531250,   9765625;
  10: 1, 55, 1375, 20625, 206250, 1443750, 7218750, 25781250, 64453125, 107421875, 107421875, 48828125;
%F Formula
The row polynomials satisfy (1+x)*P(n,x) = (1+6*x+5*x^2)*P(n-1,x) for n >= 2, with P(1,x) = 1 + 10*x + 25*x^2.
%t Mathematica
P[1, x_] := 1 + 10*x + 25*x^2;
P[n_, x_] := P[n, x] = Expand[Together[((1 + 6*x + 5*x^2)*P[n - 1, x])/(1 + x)]];
Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N)); v[1] = 1 + 10*x + 25*x^2; for(n = 2, N, my(i = n); v[i] = ((1 + 6*x + 5*x^2)*v[i - 1])/(1 + x)); 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)

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