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%I #44 Sep 29 2019 09:53:28
%S 0,2,4,32,544,15872,707584,44736512,3807514624,419730685952,
%T 58177770225664,9902996106248192,2030847773013704704,
%U 493842960380415967232,140503203207887919775744,46238368375619195682947072,17427925514250338592341622784,7458815407441059142195019251712
%N E.g.f.: -log(cos(x)*cos(x)) (even powers only).
%C Of course this is 2*log(sec(x)), so a(n) = 2*A000182(n).
%H Tewodros Amdeberhan, Victor H. Moll and Christophe Vignat, <a href="https://arxiv.org/abs/1202.1203">A probabilistic interpretation of a sequence related to Narayana Polynomials</a>, arXiv:1202.1203 [math.NT], 2012. See p. 21.
%H Tewodros Amdeberhan, Victor H. Moll and Christophe Vignat, <a href="https://web.math.rochester.edu/misc/ojac/vol8/68.pdf">A probabilistic interpretation of a sequence related to Narayana Polynomials</a>, Online Journal of Analytic Combinatorics, Issue 8, 2013. See p. 21.
%H N. J. A. Sloane, <a href="/A001469/a001469_1.pdf">Rough notes on Genocchi numbers</a>
%F G.f.: 2/Q(0) where Q(k) = 1 + x*(2*k + 1)*(2*k + 2)/( -1 + x*(2*k + 2)*(2*k + 3)/Q(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Mar 11 2013
%F G.f.: (2/G(0) - 1)*sqrt(-x), where G(k)= 2 + 2*sqrt(-x) - 4*x*(k+1)^2/G(k+1); (continued fraction). - _Sergei N. Gladkovskii_, May 29 2013
%F G.f.: 2*x*T(0), where T(k) = 1 - (k+1)*(k+2)*x/((k+1)*(k+2)*x - 1/T(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Nov 15 2013
%F a(n) ~ 2^(2*n+2) * (2*n-1)! / Pi^(2*n). - _Vaclav Kotesovec_, Feb 08 2015
%F E.g.f. (odd powers): y = 2*tan(x). - _Stanislav Sykora_, Nov 28 2016
%e G.f. = x^2+1/6*x^4+2/45*x^6+17/1260*x^8+62/14175*x^10+691/467775*x^12+...
%t nn = 20; Table[(CoefficientList[Series[-Log[Cos[x]^2], {x, 0, 2*nn}], x] * Range[0, 2*nn]!)[[n]], {n, 1, 2*nn+1, 2}] (* _Vaclav Kotesovec_, Feb 08 2015 *)
%Y Cf. A000182.
%K nonn
%O 0,2
%A Patrick Demichel (patrick.demichel(AT)hp.com)
%E Corrected by _D. S. McNeil_ and _N. J. A. Sloane_, Dec 17 2011 (The signs were wrong and the initial term was missing)