A142461 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(1 + 4\,x + 5\,x^{2} + 2\,x^{3}\right)\,P_{n-1}(x) + \left(6\,x + 18\,x^{2} + 18\,x^{3} + 6\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1123
21161633
31114336224675
4180052808960448019521
51560275040244160291200116480732483
6139216102033656896001216320011182080372736033821793
71274514135780961223871044178720006628876804957388801416396801854377955
Technical details
Node id
n-552c7386b5928d1b · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-3980a5ccbbfbe4b6

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): no

Matches ?

No OEIS matches.

Submission text ?

Atlas rows n = 0, …, ascending powers of x
%e Example
Triangle begins:
  0: 1;
  1: 1,        2;
  2: 1,       16,         16;
  3: 1,      114,        336,         224;
  4: 1,      800,       5280,        8960,         4480;
  5: 1,     5602,      75040,      244160,       291200,        116480;
  6: 1,    39216,    1020336,     5689600,     12163200,      11182080,       3727360;
  7: 1,   274514,   13578096,   122387104,    417872000,     662887680,     495738880,     141639680;
  8: 1,  1921600,  178711360,  2515448320,  12894542080,   31414190080,   39554744320,   24928583680,    6232145920;
  9: 1, 13451202, 2338620480, 50295477120, 372672518400, 1309097986560, 2468779622400, 2575009382400, 1402232832000, 311607296000;
%F Formula
The row polynomials satisfy (1+2*x+x^2)*P(n,x) = (1+4*x+5*x^2+2*x^3)*P(n-1,x) + (6*x+18*x^2+18*x^3+6*x^4)*d/dx 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[Together[((1 + 4*x + 5*x^2 + 2*x^3)*P[n - 1, x] + (6*x + 18*x^2 + 18*x^3 + 6*x^4)*D[P[n - 1, x], x])/(1 + 2*x + x^2)]];
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 + 4*x + 5*x^2 + 2*x^3)*v[i - 1] + (6*x + 18*x^2 + 18*x^3 + 6*x^4)*deriv(v[i - 1], x))/(1 + 2*x + x^2)); 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. [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