A163747 Expansion of e.g.f. 2*exp(x)*(1-exp(x))/(1+exp(2*x)).

0, -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 03 2009

Keywords

• sign

Comments

The real part of the exponential expansion of 2*((1+i)/(1+i*exp(z))-1) = (-1-i)*z + (-1/2+i/2)*z^2 + (1/3+i/3)*z^3 + (5/24-5i/24)*z^4 + (-2/15-2i/15)*z^5 + ... where i is the imaginary unit.

From Paul Curtz, Mar 12 2013: (Start)

a(n) is an autosequence of the first kind; a(n) and successive differences are:

0, -1, -1, 2, 5, -16, -61;

-1, 0, 3, 3, -21, -45, 333;

1, 3, 0, -24, -24, 378, 780;

2, -3, -24, 0, 402, 402, -11214;

-5, -21, 24, 402, 0, -11616, -11616;

-16, 45, 378, -402, -11616, 0, 514608;

61, 333, -780, -11214, 11616, 514608, 0;

The main diagonal is A000004. The inverse binomial transform is the signed sequence.

The first two upper diagonals are A002832 (median Euler numbers) signed.

Sum of the antidiagonals: 0, -2, 0, 10, 0, ... = 2*A122045(n+1) (End)

Links

• Robert Israel, Table of n, a(n) for n = 0..485
• Toufik Mansour, Howard Skogman, and Rebecca Smith, Passing through a stack k times with reversals, arXiv:1808.04199 [math.CO], 2018.
• A. Randrianarivony and J. Zeng, Une famille de polynomes qui interpole plusieurs suites classiques de nombres, Adv. Appl. Math. 17 (1996), 1-26. See Section 6 (negative of the zeroth column of matrix a_{n,k} on p. 18).

Crossrefs

Variant: A163982.

Cf. A000004, A000111, A002832, A122045.

Minus the zeroth column of A323833.

Programs

Maple

A163747 := proc(n) exp(t)*(1-exp(t))/(1+exp(2*t)) ; coeftayl(%,t=0,n) ; 2*%*n! ; end proc: # R. J. Mathar, Sep 11 2011
seq((euler(n) - 2^n*(2*euler(n,1)+euler(n,3/2)))/2 + 1, n=0..30); # Robert Israel, May 24 2016
egf := (2 - 2*I)/(exp(-x) + I); ser := series(egf, x, 24):
seq(n!*Re(coeff(ser, x, n)), n = 0..22); # Peter Luschny, Aug 09 2021

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 = -x/g[0]; CoefficientList[Series[gf, {x, 0, max}], x] (* Vaclav Kotesovec, Jan 22 2015, after Sergei N. Gladkovskii *)
Table[(EulerE[n] - 2^n (2 EulerE[n, 1] + EulerE[n, 3/2]))/2 + 1, {n, 0, 20}] (* Benedict W. J. Irwin, May 24 2016 *)

Formula

G.f.: -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

a(n) ~ n! * (cos(Pi*n/2) - sin(Pi*n/2)) * 2^(n+2) / Pi^(n+1). - Vaclav Kotesovec, Apr 23 2015

a(n) = (A122045(n) - 2^n(2*Euler(n,1) + Euler(n,3/2)))/2 + 1, where Euler(n,x) is the n-th Euler polynomial. - Benedict W. J. Irwin, May 24 2016

a(n) = 2*4^n*(HurwitzZeta(-n, 1/4) - HurwitzZeta(-n, 3/4)) + HurwitzZeta(-n, 1)*(4^(n+1) - 2^(n+1)). - Peter Luschny, Jul 21 2020

a(n) = 2^n*(Euler(n, 1/2) - Euler(n, 1)). - Peter Luschny, Mar 19 2021

a(n) = ((-2)^(n + 1)*(1 - 2^(n + 1))*Bernoulli(n + 1))/(n + 1) + Euler(n). - Peter Luschny, May 06 2021

a(n) = n!*Re([x^n]((2 - 2*i)/(i + exp(-x)))). - Peter Luschny, Aug 09 2021