%I M3066 #42 Jul 14 2024 16:15:12
%S 1,1,3,19,217,3961,105963,3908059,190065457,11785687921,907546301523,
%T 84965187064099,9504085749177097,1251854782837499881,
%U 191781185418766714683,33810804270120276636139,6796689405759438360407137,1545327493049348356667631841
%N Salié numbers.
%C There is another sequence called Salié numbers, A000795. - _Benedict W. J. Irwin_, Feb 10 2016
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 87, Problem 32.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, <a href="/A005647/b005647.txt">Table of n, a(n) for n = 0..100</a>
%H P. Bala, <a href="/A005647/a005647.pdf">A triangle for calculating A005647</a>
%H L. Carlitz, <a href="https://doi.org/10.1007/BF01298317">The coefficients of cosh x/ cos x</a>, Monatshefte für Mathematik 69(2) (1965), 129-135.
%F a(n) = A000795(n)/2^n.
%F Expand cosh x / cos x and multiply coefficients by n!/(2^(n/2)).
%F a(n) = 2^(-n)*Sum_{k=0..n} A000364(k)*binomial(2*n, 2*k)). - _Philippe Deléham_, Jul 30 2003
%F a(n) ~ (2*n)! * 2^(n+2) * cosh(Pi/2) / Pi^(2*n+1). - _Vaclav Kotesovec_, Mar 08 2014
%F G.f.: A(x) = 1/(1 - x/(1 - 2x/(1 - 5x/(1 - 8x/(1 - 13x/(1 - 18x/(1 -...))))))), a continued fraction where the coefficients are A000982 (ceiling(n^2/2)). - _Benedict W. J. Irwin_, Feb 10 2016
%t nmax = 17; se = Series[ Cosh[x]/Cos[x], {x, 0, 2*nmax}]; a[n_] := Coefficient[se, x, 2*n]*(2*n)!/2^n; Table[a[n], {n, 0, nmax}](* _Jean-François Alcover_, May 11 2012 *)
%t Join[{1},Table[SeriesCoefficient[Series[1/(1+ContinuedFractionK[Floor[(k^2+ 1)/2]*x*-1,1,{k,1,20}]),{x,0,20}],n],{n,1,20}]](* _Benedict W. J. Irwin_, Feb 10 2016 *)
%Y Cf. A000795, A000982.
%K nonn,easy,nice
%O 0,3
%A _Simon Plouffe_, _N. J. A. Sloane_