cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A321965 a(n) = n! [x^n] exp((1/(x - 1)^2 - 1)/2)/(1 - x).

Original entry on oeis.org

1, 2, 8, 46, 338, 2996, 30952, 364148, 4797116, 69854968, 1113018176, 19244304872, 358608737368, 7160626365296, 152458303437728, 3446434090192816, 82412163484132112, 2077739630757428768, 55068742629150564736, 1530394053934299827168, 44490672191650220419616
Offset: 0

Views

Author

Peter Luschny, Dec 20 2018

Keywords

Crossrefs

Row sums of A321966.

Programs

  • Maple
    egf := exp((1/(x - 1)^2 - 1)/2)/(1 - x): ser := series(egf, x, 22):
    seq(n!*coeff(ser, x, n), n=0..20);
  • Mathematica
    CoefficientList[Exp[(1/(x - 1)^2 - 1)/2]/(1 - x) + O[x]^21, x] Range[0, 20]! (* Jean-François Alcover, Jan 01 2019 *)

Formula

a(n + 3) = (n + 1)^2*(n + 2)*a(n) - (5 + 3*n)*(n + 2)*a(n + 1) + (8 + 3*n)*a(n + 2). - Robert Israel, Dec 20 2018
a(n) ~ exp(-1/3 + n^(1/3)/2 + 3*n^(2/3)/2 - n) * n^(n + 1/6) / sqrt(3). - Vaclav Kotesovec, Dec 20 2018

A322944 Coefficients of a family of orthogonal polynomials. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 6, 1, 6, 38, 15, 1, 24, 272, 188, 28, 1, 120, 2200, 2340, 580, 45, 1, 720, 19920, 30280, 11040, 1390, 66, 1, 5040, 199920, 413560, 206920, 37450, 2842, 91, 1, 40320, 2204160, 5989760, 3931200, 955920, 102816, 5208, 120, 1
Offset: 0

Views

Author

Peter Luschny, Jan 02 2019

Keywords

Comments

The polynomials represent a family of orthogonal polynomials which obey a recurrence of the form p(n, x) = (x+r(n))*p(n-1, x) - s(n)*p(n-2, x) + t(n)*p(n-3, x) - u(n)*p(n-4, x). For the details see the Maple program.
We conjecture that the polynomials have only negative and simple real roots.

Examples

			Triangle starts:
[0]    1;
[1]    1,      1;
[2]    2,      6,      1;
[3]    6,     38,     15,      1;
[4]   24,    272,    188,     28,     1;
[5]  120,   2200,   2340,    580,    45,    1;
[6]  720,  19920,  30280,  11040,  1390,   66,  1;
[7] 5040, 199920, 413560, 206920, 37450, 2842, 91, 1;
Production matrix starts:
   1;
   1,    1;
   3,    5,    1;
   6,   18,    9,    1;
   6,   42,   45,   13,    1;
   0,   48,  132,   84,   17,    1;
   0,    0,  180,  300,  135,   21,    1;
   0,    0,    0,  480,  570,  198,   25,    1;
		

Crossrefs

p(n, 1) = A322943(n) (row sums); p(n, 0) = n! = A000142(n).
A321966 (m=2), this sequence (m=3).
Cf. A321620.

Programs

  • Maple
    P := proc(n) option remember; local a, b, c, d;
    a := n -> 4*n-3; b := n -> 3*(n-1)*(2*n-3);
    c := n -> (n-1)*(n-2)*(4*n-9); d := n -> (n-2)*(n-1)*(n-3)^2;
    if n = 0 then return 1 fi;
    if n = 1 then return x + 1 fi;
    if n = 2 then return x^2 + 6*x + 2 fi;
    if n = 3 then return x^3 + 15*x^2 + 38*x + 6 fi;
    expand((x+a(n))*P(n-1) - b(n)*P(n-2) + c(n)*P(n-3) - d(n)*P(n-4)) end:
    seq(print(P(n)), n=0..9); # Computes the polynomials.
  • Mathematica
    a[n_] := 4n - 3;
    b[n_] := 3(n - 1)(2n - 3);
    c[n_] := (n - 1)(n - 2)(4n - 9);
    d[n_] := (n - 2)(n - 1)(n - 3)^2;
    P[n_] := P[n] = Switch[n, 0, 1, 1, x + 1, 2, x^2 + 6x + 2, 3, x^3 + 15x^2 + 38x + 6, _, Expand[(x + a[n]) P[n - 1] - b[n] P[n - 2] + c[n] P[n - 3] - d[n] P[n - 4]]];
    Table[CoefficientList[P[n], x], {n, 0, 9}] (* Jean-François Alcover, Jun 15 2019, from Maple *)
  • Sage
    # uses[riordan_square from A321620]
    R = riordan_square((1 - 3*x)^(-1/3), 9, True).inverse()
    for n in (0..8): print([(-1)^(n-k)*c for (k, c) in enumerate(R.row(n)[:n+1])])

Formula

Let R be the inverse of the Riordan square [see A321620] of (1 - 3*x)^(-1/3) then T(n, k) = (-1)^(n-k)*R(n, k).
Showing 1-2 of 2 results.