Real-rooted atlas

A142462 → x ↦ x/(x+1)

nonnegative coefficients

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

Recurrence

\(\left(1 + 2\,x + x^{2}\right)\,P_{n}(x) = \left(1 + 4\,x + 5\,x^{2} + 2\,x^{3}\right)\,P_{n-1}(x) + \left(7\,x + 21\,x^{2} + 21\,x^{3} + 7\,x^{4}\right)\,P_{n-1}'(x)\)
\(P_{0}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7Pn(1)
011
1123
21181837
31146432288867
411170779413248662428837
5193621274404161604968001987201248483
61748981995858111945602397888022057920735264066654757
7159918630611952278214048952862400151345152011323065603235161604231561827

All 15 rows

Technical details
Node id
n-58a84645df5825fb · revision 0 · JSON
Definition source
derived:mobius_x_over_x_plus_one
Fingerprint
f6-d4f1df9ae9c004ef

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,       18,         18;
  3: 1,      146,        432,          288;
  4: 1,     1170,       7794,        13248,          6624;
  5: 1,     9362,     127440,       416160,        496800,        198720;
  6: 1,    74898,    1995858,     11194560,      23978880,      22057920,       7352640;
  7: 1,   599186,   30611952,    278214048,     952862400,    1513451520,    1132306560,     323516160;
  8: 1,  4793490,  464571954,   6610500288,   34031932704,   83070126720,  104686888320,   65997296640,   16499324160;
  9: 1, 38347922, 7011720720, 152864157600, 1138967555040, 4011482543040, 7575130886400, 7906087918080, 4306323605760, 956960801280;
%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) + (7*x+21*x^2+21*x^3+7*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] + (7*x + 21*x^2 + 21*x^3 + 7*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] + (7*x + 21*x^2 + 21*x^3 + 7*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