%I #42 Jun 07 2020 18:43:03
%S 1,3,12,63,432,3798,41472,543483,8301312,144502218,2818685952,
%T 60826110678,1437615931392,36914181252588,1022923413061632,
%U 30419533530730323,966125479213596672,32634383834158752258,1168128785923721920512
%N The case S(2,0,-1) of the family of self-convolutive recurrences studied by Martin and Kearney.
%H G. C. Greubel, <a href="/A172450/b172450.txt">Table of n, a(n) for n = 1..400</a>
%H A. V. Belitsky, G. P. Korchemsky, <a href="https://arxiv.org/abs/2003.01121">Octagon at finite coupling</a>, arXiv:2003.01121 [hep-th], 2020.
%H R. J. Martin and M. J. Kearney, <a href="https://dx.doi.org/10.1007/s00010-010-0051-0">An exactly solvable self-convolutive recurrence</a>, Aequat. Math., 80 (2010), 291-318. see p. 294.
%H R. J. Martin and M. J. Kearney, <a href="http://arXiv.org/abs/1103.4936">An exactly solvable self-convolutive recurrence</a>, arXiv:1103.4936 [math.CO], 2011.
%F a(n) = 2 * n * a(n-1) - Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - _Michael Somos_, Jul 24 2011
%F G.f.: x / (1 - 3*x / (1 - 1*x / (1 - 5*x / (1 - 3*x / (1 - 7*x / (1 - 5*x / (1 - 9*x / (1 - 7*x / (1 - 11*x / (1 - 9*x / (1 - 13*x / (1 - 11*x / ... )))))))))))). - _Michael Somos_, Jan 03 2013
%F G.f.: (1/(2*Q(0) - 1) - 1)/(2*x) where Q(k) = 1 - x*(2*k+1)/( 1 - x*(2*k+1)/Q(k+1) ); (continued fraction ). - _Sergei N. Gladkovskii_, Apr 02 2013
%F G.f.: Q(0)/x - 1/x, where Q(k)= 1 - x*(2*k-1)/(1 - x*(2*k+3)/Q(k+1)); (continued fraction). - _Sergei N. Gladkovskii_, May 21 2013
%F a(n) ~ 2^n * (n-1)! / Pi. - _Vaclav Kotesovec_, Jan 19 2015
%F a(n) = (1/2)*4^n*[x^n](BesselK(0,-1/x)/BesselK(1,-1/x) - 1). - _Peter Luschny_, Dec 11 2017
%e x + 3*x^2 + 12*x^3 + 63*x^4 + 432*x^5 + 3798*x^6 + 41472*x^7 + 543483*x^8 + ...
%p ser := series(BesselK(0,-1/x)/BesselK(1,-1/x) - 1, x, 20):
%p seq((1/2)*4^n*coeff(ser,x,n), n=0..19); # _Peter Luschny_, Dec 11 2017
%t Clear[a]; a[1] = 1; a[n_]:= a[n] = 2*n*a[n-1] - Sum[a[k]*a[n-k], {k, 1, n-1}]; Table[a[n], {n, 1, 20}] (* _Vaclav Kotesovec_, Jan 19 2015 *)
%o (PARI) {a(n) = local(A); if( n<1, 0, A = vector(n); A[1] = 1; for( k=2, n, A[k] = 2 * k * A[k-1] - sum( j=1, k-1, A[j] * A[k-j])); A[n])} /* _Michael Somos_, Jul 24 2011 */
%Y Cf. A005411.
%K nonn
%O 1,2
%A _N. J. A. Sloane_, Nov 20 2010