A119880 Expansion of e.g.f. exp(2x)*sech(x).

1, 2, 3, 2, -3, 2, 63, 2, -1383, 2, 50523, 2, -2702763, 2, 199360983, 2, -19391512143, 2, 2404879675443, 2, -370371188237523, 2, 69348874393137903, 2, -15514534163557086903, 2, 4087072509293123892363, 2, -1252259641403629865468283, 2, 441543893249023104553682823

Offset: 0

Author

Paul Barry, May 26 2006

Keywords

• easy
• sign

Comments

Transform of 2^n under the matrix A119879.

Also the Swiss-Knife polynomials A153641 evaluated at x=2. - Peter Luschny, Nov 23 2012

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Paul Barry, Eulerian-Dowling Polynomials as Moments, Using Riordan Arrays, arXiv:1702.04007 [math.CO], 2017.

Crossrefs

Cf. A119879, A119880, A122045.

Programs

Magma

EulerPoly:= func< n,x | (&+[ (&+[ (-1)^j*Binomial(k,j)*(x+j)^n : j in [0..k]])/2^k: k in [0..n]]) >;
A119880:= func< n | (-2)^n*EulerPoly(n, -1/2) >;
[A119880(n): n in [0..40]]; // G. C. Greubel, Jun 07 2023

Maple

A119880_list := proc(n) local S,A,m,k;
A := array(0..n-1,0..n-1); S := NULL;
for m from 0 to n-1 do
   A[m,0] := (-2)^m*euler(m,0);
   for k from m-1 by -1 to 0 do
       A[k,m-k] := A[k+1,m-k-1] + A[k,m-k-1] od;
    S := S,A[0,m] od;
S end:
A119880_list(31); # Peter Luschny, Jun 15 2012
P := proc(n,x) option remember; if n = 0 then 1 else
  (n*x-(1/2)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x);
  expand(%) fi end:
A119880 := n -> (-1)^n*subs(x=-1, P(n,x)):
seq(A119880(n), n=0..30);  # Peter Luschny, Mar 07 2014

Mathematica

Table[2 (1 + Zeta[-n] (2^n - 1) + 2^(2n+1) Zeta[-n, 3/4]), {n, 0, 30}] (* Peter Luschny, Jul 16 2013 *)
Range[0, 30]! CoefficientList[Series[Exp[2 x] Sech[x], {x, 0, 30}], x] (* Vincenzo Librandi, Mar 08 2014 *)

Sage

def skp(n, x):
    A = lambda k: 0 if (k+1)%4 == 0 else (-1)^((k+1)//4)*2^(-(k//2))
    return add(A(k)*add((-1)^v*binomial(k,v)*(v+x+1)^n for v in (0..k)) for k in (0..n))
A119880 = lambda n: skp(n,2)
[A119880(n) for n in (0..30)]  # Peter Luschny, Nov 23 2012

Formula

a(n) = Sum_{k=0..n} A119879(n,k) * 2^k.

From Sergei N. Gladkovskii, Oct 14 2012 to Dec 16 2013: (Start)

Continued fractions:

G.f.: 1/U(0) where U(k) = 1 - x - x*(k+1)/(1 + x*(k+1)/U(k+1)).

G.f.: 1/Q(0), where Q(k) = 1 - 3*x + x*(k+1)/(1-x*(k+1)/Q(k+1)).

G.f.: x/(1-x)/Q(0) + 1/(1-x), where Q(k) = 1 - x + x^2*(k+1)*(k+2)/Q(k+1).

G.f.: T(0)/(1-2*x), where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 + (1-2*x)^2/T(k+1)).

E.g.f.: 2/Q(0), where Q(k) = 1 + 3^k/( 1 - x/( x - 3^k*(k+1)/Q(k+1))). (End)

a(n) = 2*(1+zeta(-n)*(2^n-1)+2^(2*n+1)*zeta(-n,3/4)). - Peter Luschny, Jul 16 2013

a(n) = (-2)^n*Euler(n, -1/2). - Peter Luschny, Jul 21 2020