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.

Showing 1-3 of 3 results.

A194649 Triangle of coefficients of a sequence of polynomials related to the enumeration of linear labeled rooted trees.

Original entry on oeis.org

1, 1, 3, 4, 13, 36, 24, 75, 316, 432, 192, 541, 3060, 6360, 5760, 1920, 4683, 33244, 92880, 127680, 86400, 23040, 47293, 403956, 1418424, 2620800, 2688000, 1451520, 322560, 545835, 5449756, 23051952, 53548992, 73785600, 60318720, 27095040, 5160960, 7087261, 80985780, 400813080, 1122145920, 1943867520, 2133734400
Offset: 0

Views

Author

Peter Bala, Sep 01 2011

Keywords

Comments

Define the sequence of polynomials {P(n,x)}n>=0 recursively by setting P(0,x) = 1, P(1,x) = 1 and P(n+1,x) = d/dx((1+x)*(1+2*x)*P(n,x)) for n >= 1. The first few values are P(2,x) = 3 + 4*x, P(3,x) = 13 + 36*x + 24*x^2 and P(4,x) = 75 + 316*x + 432*x^2 + 192*x^3.
This triangle shows the coefficients of the P(n,x) in ascending powers of x. The values of P(n,x) at an integer or half-integer value of x enumerate linear labeled rooted trees: in particular we have P(n,0) = A000670(n), P(n,1/2) = A050351(n), P(n,1) = A050352(n) and P(n,3/2) = A050353(n).
More generally, for m >= 2, P(n,m/2-1), n = 0,1,2,... counts m level linear labeled rooted trees (see the e.g.f. below and the comment of Benoit Cloitre in A050351).

Examples

			Triangle begins
n\k|......0.......1........2........3........4........5.......6
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
.0.|......1
.1.|......1
.2.|......3.......4
.3.|.....13......36.......24
.4.|.....75.....316......432......192
.5.|....541....3060.....6360.....5760.....1920
.6.|...4683...33244....92880...127680....86400....23040
.7.|..47293..403956..1418424..2620800..2688000..1451520..322560
..

Links

Crossrefs

Cf. A000670, A002866 (main diagonal), A050351, A050352, A050353, A083411 (1/4*column 1).

Programs

  • Mathematica
    T[0, 0] = T[1, 0] = 1; T[n_, k_] /; 0 <= k <= n-1 := T[n, k] = (k+1)*(2* T[n-1, k-1] + 3*T[n-1, k] + T[n-1, k+1]); T[, ] = 0;
{1}~Join~Table[T[n, k], {n, 1, 9}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Nov 13 2019 *)

Formula

T(n,k) = 2^k*Sum_{i = k+1..n} Stirling2(n,i)*i!*binomial(i-1,k).
Recurrence: T(n+1,k) = (k+1)*(2*T(n,k-1)+3*T(n,k)+T(n,k+1)).
E.g.f.: G(x,t) := 1 + (1-exp(t))/((2*x+1)*exp(t)-2*x-2) = Sum_{n>=0} P(n,x)*t^n/n! = 1 + t + (3 + 4*x)*t^2/2! + (13 + 36*x + 24*x^2)*t^3/3! + ....
Column k generating function: 2^k*((exp(x)-1)/(2-exp(x)))^(k+1) (apart from initial term 1 when k = 0).
The generating function G(x,t) satisfies the partial differential equation d/dx((1+x)*(1+2*x)*G(x,t)) - d/dt(G(x,t)) = 2*(2x+1). Hence the row polynomials P(n,x) satisfy the defining recurrence P(n+1,x) = d/dx((1+x)*(2+x)*P(n,x)), with P(0,x) = P(1,x) = 1.
Reflection property: P(n,x) = (-1)^n*P(n,-x-3/2).
The polynomial P(n,x) has all real zeros, lying in the interval [-1,-1/2] (apply [Liu et al, Theorem 1.1, Corollary 1.2] with f(x) = P(n,x-1/2) and g(x) = P'(n,x-1/2) and use the reflection property).
Row sums are A050352; Column 0: A000670; Column 1: 4*A083411; Main diagonal: A002866.

A083384 a(n) = n*Sum(((k-1)/2)*k!*Stirling_2(n,k),k=1..n).

Original entry on oeis.org

0, 2, 27, 316, 3825, 49866, 706923, 10899512, 182218005, 3289724710, 63865092159, 1327750936788, 29447495757225, 694257067232834, 17343019158929235, 457695211932767344, 12726295039220109885, 371902424983010438238, 11396594412860395106151, 365458808048854606362380
Offset: 1

Views

Author

N. J. A. Sloane, Jun 07 2003

Keywords

Crossrefs

Cf. A000670, A083385, A083411, A083410.

Programs

  • Magma
    [n*&+[(k-1)/2*Factorial(k)*StirlingSecond(n, k): k in [0..n]]: n in [1..25]]; //  Vincenzo Librandi, Sep 01 2018
  • Mathematica
    a[n_] := n Sum[1/2 (k-1) k! StirlingS2[n, k], {k, 1, n}];
Array[a, 20] (* Jean-François Alcover, Sep 01 2018 *)
Rest[Range[0, 19]! CoefficientList[Series[x (Exp[x] - 1) Exp[x] / (2 - Exp[x])^3, {x, 0, 19}], x]] (* Vincenzo Librandi, Sep 01 2018 *)

Formula

Equals A083385(n) - n*A000670(n).
E.g.f.: x*(exp(x)-1)*exp(x)/(2-exp(x))^3. - Vladeta Jovovic, Sep 14 2003
a(n) ~ n! * n^2 / (8 * (log(2))^(n+2)). - Vaclav Kotesovec, Feb 18 2017

A344054 a(n) = Sum_{k = 0..n} E1(n, k)*k^2, where E1 are the Eulerian numbers A173018.

Original entry on oeis.org

0, 0, 1, 8, 64, 540, 4920, 48720, 524160, 6108480, 76809600, 1037836800, 15008716800, 231437606400, 3792255667200, 65819609856000, 1206547550208000, 23297526540288000, 472708591939584000, 10055994967130112000, 223826984752250880000, 5202760944485744640000, 126075414965721661440000, 3179798058882852126720000, 83346901966165164687360000, 2267221868000212451328000000
Offset: 0

Views

Author

Peter Luschny, May 11 2021

Keywords

Comments

The Eulerian transform of the squares.

Crossrefs

Transforms of the squares: A151881 (StirlingCycle), A033452 (StirlingSet), A105219 (Laguerre), A103194 (Lah), A065096 (SchröderBig), A083411 (Fubini), A141222 (Narayana), A000330 (Units A000012).
Cf. A173018.

Programs

  • Maple
    a := n -> add(combinat[eulerian1](n, k)*k^2, k = 0..n):
# Recurrence:
a := proc(n) option remember; if n < 2 then 0 elif n = 2 then 1 else
((n-3)*(n-1)*(23*n-44)*a(n-2) + ((159 - 7*n)*n - 286)*a(n-1))/(16*(n - 2)) fi end:
seq(a(n), n = 0..29);
  • Mathematica
    a[n_] := Sum[Sum[(-1)^j Binomial[n + 1, j] k^2 (k + 1 - j)^n, {j,0,k}], {k,0,n}]; a[0] := 0; Table[a[n], {n, 0, 25}]
  • SageMath
    def aList(len):
    R. = PowerSeriesRing(QQ, default_prec=len+2)
    f = x^2*(-x^2 + x - 3)/(6*(x - 1)^3)
    return f.egf_to_ogf().list()[:len]
print(aList(20))

Formula

a(n) = n! * [x^n] x^2*(-x^2 + x - 3)/(6*(x - 1)^3).
a(n) = Sum_{k=0..n} Sum_{j=0..k} (-1)^j*binomial(n + 1, j)*k^2*(k + 1 - j)^n.
a(n) = ((n - 3)*(n - 1)*(23*n - 44)*a(n-2) + ((159 - 7*n)*n - 286)*a(n-1))/(16*(n - 2)) for n >= 3.
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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