A163982 Real part of the coefficient [x^n] of the expansion of (1+i)/(1-i*exp(x)) - 1 multiplied by 2*n!, where i is the imaginary unit.

-2, -1, 1, 2, -5, -16, 61, 272, -1385, -7936, 50521, 353792, -2702765, -22368256, 199360981, 1903757312, -19391512145, -209865342976, 2404879675441, 29088885112832, -370371188237525, -4951498053124096, 69348874393137901

Offset: 0

Author

Roger L. Bagula, Aug 07 2009

Keywords

• sign

Comments

The sequence is a signed variant of A163747 and starts with a two instead of a zero.

From Paul Curtz, Mar 20 2013: (Start)

-a(n) and successive differences are:

2, 1, -1, -2, 5, 16, -61, -272;

-1, -2, -1, 7, 11, -77, -211, 1657, ...

-1, 1, 8, 4, -88, -134, 1868, 4894, ...

2, 7, -4, -92, -46, -46, 2002, 3026, ...

5, -11, -88, 46, 2048, 1024, -72928, ...

-16, -77, 134, 2002, -1024, -73952, -36976, ...

-61, 211, 1868, -3026, -72928, ...

272, 1657, -4894, -69902, ...

This is an autosequence: The inverse binomial transform (left column of the array of differences) is the signed sequence. The main diagonal 2, -2, 8, -92, ... doubles the entries of the first upper diagonal 1, -1, 4, -46, ... = A099023(n).

Sum of the antidiagonals: 2, 0, -4, 0, 32, ... = 2*A155585(n+1). (End)

Links

• G. C. Greubel, Table of n, a(n) for n = 0..480
• Toufik Mansour, Howard Skogman, Rebecca Smith, Passing through a stack k times with reversals, arXiv:1808.04199 [math.CO], 2018.

Crossrefs

Cf. A163747.

Programs

Maple

A163982 := n -> -2^n*(euler(n,1/2)+euler(n,1)): # Peter Luschny, Nov 25 2010
A163982 := proc(n)
    (1+I)/(1-I*exp(x))-1 ;
    coeftayl(%,x=0,n) ;
    Re(%*2*n!) ;
end proc; # R. J. Mathar, Mar 26 2013

Mathematica

f[t_] = (1 + I)/(1 - I*Exp[t]) - 1; Table[Re[2*n!*SeriesCoefficient[Series[f[t], {t, 0, 30}], n]], {n, 0, 30}]
max = 20; Clear[g]; g[max + 2] = 1; g[k_] := g[k] = 1 + x + (4*k+3)*(k+1)*x^2 /( 1 + (4*k+5)*(k+1)*x^2 / g[k+1]); gf = -2 - x/g[0]; CoefficientList[Series[gf, {x, 0, max}], x] (* Vaclav Kotesovec, Jan 22 2015, after Sergei N. Gladkovskii *)
With[{nn = 50}, CoefficientList[Series[(-2)*Exp[t/2]*Cosh[t/2]/Cosh[t], {t, 0, nn}], t]*Range[0, nn]!] (* G. C. Greubel, Aug 24 2017 *)

PARI

t='t+O('t^10); Vec(serlaplace((-2)*exp(x/2)*cosh(x/2)/cosh(x))) \\ G. C. Greubel, Aug 24 2017

Formula

Let ((1 + i)/(1 - i*exp(t)) - 1) = a(n) + I*b(n); abs(a(n)) = abs(b(n)).

a(n) = -2^n*(E_{n}(1/2) + E_{n}(1)), E_{n}(x) Euler polynomial. - Peter Luschny, Nov 25 2010

E.g.f.: -(1/cosh(x) + tanh(x)) - 1. - Sergei N. Gladkovskii, Dec 11 2013

G.f.: -2 - x/W(0), where W(k) = 1 + x + (4*k+3)*(k+1)*x^2 /( 1 + (4*k+5)*(k+1)*x^2 /W(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Jan 22 2015

E.g.f.: (-2)*exp(x/2)*cosh(x/2)/cosh(x). - G. C. Greubel, Aug 24 2017