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%I #34 Nov 07 2018 14:44:15
%S -2,-1,1,2,-5,-16,61,272,-1385,-7936,50521,353792,-2702765,-22368256,
%T 199360981,1903757312,-19391512145,-209865342976,2404879675441,
%U 29088885112832,-370371188237525,-4951498053124096,69348874393137901
%N 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.
%C The sequence is a signed variant of A163747 and starts with a two instead of a zero.
%C From _Paul Curtz_, Mar 20 2013: (Start)
%C -a(n) and successive differences are:
%C 2, 1, -1, -2, 5, 16, -61, -272;
%C -1, -2, -1, 7, 11, -77, -211, 1657, ...
%C -1, 1, 8, 4, -88, -134, 1868, 4894, ...
%C 2, 7, -4, -92, -46, -46, 2002, 3026, ...
%C 5, -11, -88, 46, 2048, 1024, -72928, ...
%C -16, -77, 134, 2002, -1024, -73952, -36976, ...
%C -61, 211, 1868, -3026, -72928, ...
%C 272, 1657, -4894, -69902, ...
%C 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).
%C Sum of the antidiagonals: 2, 0, -4, 0, 32, ... = 2*A155585(n+1). (End)
%H G. C. Greubel, <a href="/A163982/b163982.txt">Table of n, a(n) for n = 0..480</a>
%H Toufik Mansour, Howard Skogman, Rebecca Smith, <a href="https://arxiv.org/abs/1808.04199">Passing through a stack k times with reversals</a>, arXiv:1808.04199 [math.CO], 2018.
%F Let ((1 + i)/(1 - i*exp(t)) - 1) = a(n) + I*b(n); abs(a(n)) = abs(b(n)).
%F a(n) = -2^n*(E_{n}(1/2) + E_{n}(1)), E_{n}(x) Euler polynomial. - _Peter Luschny_, Nov 25 2010
%F E.g.f.: -(1/cosh(x) + tanh(x)) - 1. - _Sergei N. Gladkovskii_, Dec 11 2013
%F 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
%F E.g.f.: (-2)*exp(x/2)*cosh(x/2)/cosh(x). - _G. C. Greubel_, Aug 24 2017
%p A163982 := n -> -2^n*(euler(n,1/2)+euler(n,1)): # _Peter Luschny_, Nov 25 2010
%p A163982 := proc(n)
%p (1+I)/(1-I*exp(x))-1 ;
%p coeftayl(%,x=0,n) ;
%p Re(%*2*n!) ;
%p end proc; # _R. J. Mathar_, Mar 26 2013
%t f[t_] = (1 + I)/(1 - I*Exp[t]) - 1; Table[Re[2*n!*SeriesCoefficient[Series[f[t], {t, 0, 30}], n]], {n, 0, 30}]
%t 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_ *)
%t 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 *)
%o (PARI) t='t+O('t^10); Vec(serlaplace((-2)*exp(x/2)*cosh(x/2)/cosh(x))) \\ _G. C. Greubel_, Aug 24 2017
%Y Cf. A163747.
%K sign
%O 0,1
%A _Roger L. Bagula_, Aug 07 2009
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