A078812 → x ↦ x+1 → reverse

nonnegative coefficients

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

Recurrence

\(\left(-2 + 2\,n\right)\,P_{n}(x) = \left(-2 - 12\,x - 10\,x^{2} + 2\,n + 9\,n x + 5\,n x^{2}\right)\,P_{n-1}(x) + \left(-x - 6\,x^{2} - 5\,x^{3}\right)\,P_{n-1}'(x)\)
\(P_{1}(x) = 1\)

Rows ?

nx0x1x2x3x4x5x6x7x8x9x10x11x12x13x14Pn(1)
111
2134
316815
419252156
5112519055209
611586234300144780
71181304809519543772911
812118385523053573293998710864
91242451386474010008127078850258440545
1012731621008715233734042643398261956765151316
11130396302414770481321058051546681436827650017711564719
1213348541852352690342241486448710567260464148221016463682107560
1313658356103568515789649774211250361807585201126014697786329161213937865521
1413969073265203026076694790425279204941207698530569362984578102179912531781129354524
151428069360734254112461694517521373612027225206838902608462823375736140636535082270832040109552575
Technical details
Node id
n-e9f75ddaed25275f · revision 0 · JSON
Definition source
derived:reversal
Fingerprint
f6-05007905755e7549

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 -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): no

Matches ?

No OEIS matches.

Submission text ?

Atlas rows n = 1, …, ascending powers of x
%e Example
Triangle begins:
   1: 1;
   2: 1,  3;
   3: 1,  6,   8;
   4: 1,  9,  25,   21;
   5: 1, 12,  51,   90,   55;
   6: 1, 15,  86,  234,  300,   144;
   7: 1, 18, 130,  480,  951,   954,   377;
   8: 1, 21, 183,  855, 2305,  3573,  2939,   987;
   9: 1, 24, 245, 1386, 4740, 10008, 12707,  8850,  2584;
  10: 1, 27, 316, 2100, 8715, 23373, 40426, 43398, 26195, 6765;
%F Formula
The row polynomials satisfy (2*n-2)*P(n,x) = (2*n-2+(9*n-12)*x+(5*n-10)*x^2)*P(n-1,x) - (x+6*x^2+5*x^3)*d/dx P(n-1,x) for n >= 2, with P(1,x) = 1.
%t Mathematica
P[1, x_] := 1;
P[n_, x_] := P[n, x] = Expand[Together[((2*n - 2 + (9*n - 12)*x + (5*n - 10)*x^2)*P[n - 1, x] - (x + 6*x^2 + 5*x^3)*D[P[n - 1, x], x])/(2*n - 2)]];
Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten
%o PARI/GP
P(N) = {my(v = vector(N)); v[1] = 1; for(n = 2, N, my(i = n); v[i] = ((2*n - 2 + (9*n - 12)*x + (5*n - 10)*x^2)*v[i - 1] - (x + 6*x^2 + 5*x^3)*deriv(v[i - 1], x))/(2*n - 2)); 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

Candidate equalities ?

Same fingerprint ?

References

  1. [Fav35] Jean Favard. Sur les polynômes de tchebicheff. C. R. Acad. Sci. Paris, 200:2052–2053, 1935.
  2. [Sze39] G. Szegő. Orthogonal Polynomials. Colloquium publications, American Mathematical Society, 1939. doi:10.1090/coll/023