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A193540 E.g.f.: Pi/(sqrt(2)*L) * (1 + 2*Sum_{n>=1} cos(2*Pi*n*x/L)/cosh(n*Pi)) where L = Lemniscate constant.

%I A193540 #43 Feb 16 2025 08:33:15
%S A193540 1,-1,9,-153,4977,-261009,20039481,-2121958377,296297348193,
%T A193540 -52750142341281,11662264481073129,-3134732109393169593,
%U A193540 1006734732695870345937,-380718482718134681818929,167456229155543640166939161,-84761007600911799530893148937
%N A193540 E.g.f.: Pi/(sqrt(2)*L) * (1 + 2*Sum_{n>=1} cos(2*Pi*n*x/L)/cosh(n*Pi)) where L = Lemniscate constant.
%C A193540 L = Lemniscate constant = 2*(Pi/2)^(3/2)/gamma(3/4)^2 = 2.62205755429...
%C A193540 Compare the definition with that of the dual sequence A193543.
%H A193540 Vaclav Kotesovec, <a href="/A193540/b193540.txt">Table of n, a(n) for n = 0..200</a>
%H A193540 T. J. Stieltjes, <a href="https://archive.org/details/oeuvresthomasja02stierich/page/n189/mode/2up">LXV. Sur les dérivées de sec x</a>, p. 181, Oeuvres complètes, tome 2, Noordhoff, 1918, 617 p.
%H A193540 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RamanujanCosCoshIdentity.html">Ramanujan Cos/Cosh Identity</a>.
%F A193540 Given e.g.f. A(x), define the e.g.f. of A193543:
%F A193540 B(x) = sqrt(2)*Pi/(2*L) * (1 + 2*Sum_{n>=1} cosh(2*Pi*n*x/L) / cosh(n*Pi)),
%F A193540 then A(x)^-2 + B(x)^-2 = 2 by Ramanujan's cos/cosh identity.
%F A193540 ...
%F A193540 E.g.f. equals the reciprocal of the e.g.f. of A193541.
%F A193540 O.g.f. = 1/(1 + 1^2*x/(1 + 2*2^2*x/(1 + 3^2*x/(1 + 2*4^2*x/(1 + 5^2*x/(1 + 2*6^2*x/(1 + 7^2*x/(1 + 2*8^2*x/(1+...))))))))) (continued fraction).
%F A193540 G.f.: 1/Q(0) where p=2, Q(k) = 1 + x*(2*k+1)^2/( 1 + p*x*(2*k+2)^2/Q(k+1) ); (continued fraction due to T. J. Stieltjes). - _Sergei N. Gladkovskii_, Mar 22 2013
%F A193540 a(n) ~ (-1)^n * 2^(7*n + 4) * Pi^(n+1) * n^(2*n + 1/2) / (exp(2*n) * Gamma(1/4)^(4*n + 2)). - _Vaclav Kotesovec_, Nov 29 2020
%e A193540 E.g.f.: A(x) = 1 - x^2/2! + 9*x^4/4! - 153*x^6/6! + 4977*x^8/8! - 261009*x^10/10! + 20039481*x^12/12! +...+ a(n)*x^(2*n)/(2*n)! +...
%e A193540 where
%e A193540 A(x)*sqrt(2)*L/Pi = 1 + 2*cos(2*Pi*x/L)/cosh(Pi) + 2*cos(4*Pi*x/L)/cosh(2*Pi) + 2*cos(6*Pi*x/L)/cosh(3*Pi) +...
%e A193540 Let B(x) equal the e.g.f. of A193543, where:
%e A193540 B(x)*sqrt(2)*L/Pi = 1 + 2*cosh(2*Pi*x/L)/cosh(Pi) + 2*cosh(4*Pi*x/L)/cosh(2*Pi) + 2*cosh(6*Pi*x/L)/cosh(3*Pi) +...
%e A193540 explicitly,
%e A193540 B(x) = 1 + x^2/2! + 9*x^4/4! + 153*x^6/6! + 4977*x^8/8! + 261009*x^10/10! + 20039481*x^12/12! +...
%e A193540 then A(x)^-2 + B(x)^-2 = 2
%e A193540 as illustrated by:
%e A193540 A(x)^-2 = 1 + 2*x^2/2! - 144*x^6/6! + 96768*x^10/10! - 268240896*x^14/14! +...
%e A193540 B(x)^-2 = 1 - 2*x^2/2! + 144*x^6/6! - 96768*x^10/10! + 268240896*x^14/14! +...
%e A193540 ...
%e A193540 O.g.f.: 1 - x + 9*x^2 - 153*x^3 + 4977*x^4 - 261009*x^5 + 20039481*x^6 +...+ a(n)*x^n +...
%e A193540 O.g.f.: 1/(1 + x/(1 + 8*x/(1 + 9*x/(1 + 32*x/(1 + 25*x/(1 + 72*x/(1 + 49*x/(1 + 128*x/(1+...))))))))).
%t A193540 a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ Tan[ JacobiAmplitude[ x, -1]] / Tan[ JacobiAmplitude[ 2 x, -1] / 2], {x, 0, m}]]]; (* _Michael Somos_, Oct 18 2011 *)
%t A193540 a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ JacobiND[ x, -1], {x, 0, m}]]]; (* _Michael Somos_, Oct 18 2011 *)
%t A193540 Table[SeriesCoefficient[InverseSeries[Series[EllipticF[x, 1/2], {x, 0, 32}]], 2 n + 1] (2 n + 1)! 2^n, {n, 0, 15}] (* _Benedict W. J. Irwin_, Apr 04 2017 *)
%t A193540 Table[SeriesCoefficient[JacobiDN[Sqrt[2] x, 1/2], {x, 0, 2 k}] (2 k)!, {k, 0, 20}] (* _Jan Mangaldan_, Nov 28 2020 *)
%t A193540 nmax = 20; s = CoefficientList[Series[JacobiDN[Sqrt[2] x, 1/2], {x, 0, 2*nmax}], x] * Range[ 0, 2*nmax]!; Table[s[[2*n + 1]], {n, 0, nmax}] (* _Vaclav Kotesovec_, Nov 29 2020 *)
%o A193540 (PARI) {a(n)=local(R,L=2*(Pi/2)^(3/2)/gamma(3/4)^2);
%o A193540 R=Pi/(sqrt(2)*L)*(1 + 2*suminf(m=1,cos(2*Pi*m*x/L +O(x^(2*n+1)))/cosh(m*Pi)));
%o A193540 round((2*n)!*polcoeff(R,2*n))}
%Y A193540 Cf. A193541, A193542, A193543, A193544, A193545.
%K A193540 sign
%O A193540 0,3
%A A193540 _Paul D. Hanna_, Jul 29 2011