%I M4535 #32 Nov 06 2017 02:41:16
%S 1,8,48,264,1408,7432,39152,206600,1093760,5813000,31019568,166188552,
%T 893763840,4823997960,26124870640,141926904328,773293020928,
%U 4224773978632,23139861329456,127039971696392,698993630524032,3853860616119048,21288789223825648
%N Royal paths in a lattice.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Vincenzo Librandi, <a href="/A006321/b006321.txt">Table of n, a(n) for n = 0..1000</a>
%H G. Kreweras, <a href="http://www.numdam.org/item?id=BURO_1973__20__3_0">Sur les hiérarchies de segments</a>, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973).
%H G. Kreweras, <a href="/A001844/a001844.pdf">Sur les hiérarchies de segments</a>, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)
%F a(n) = (4/n)*sum(binomial(n, j)*binomial(n+3+j, n-1), j=0..n) (n>0). - _Emeric Deutsch_, Aug 19 2004
%F Recurrence: n*(n+4)*a(n) = (5*n^2+14*n+21)*a(n-1) + (5*n^2-4*n+12)*a(n-2) - (n-3)*(n+1)*a(n-3). - _Vaclav Kotesovec_, Oct 05 2012
%F a(n) ~ 2*sqrt(816+577*sqrt(2))*(3+2*sqrt(2))^n/(sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 05 2012
%F G.f.: (x^4-8*x^3+16*x^2-8*x+1+sqrt(x^2-6*x+1)*(x-1)*(x^2-4*x+1))/(2*x^4). - _Mark van Hoeij_, Apr 16 2013
%p 1,seq(4*sum(binomial(n,j)*binomial(n+3+j,n-1),j=0..n)/n,n=1..17);
%t Flatten[{1, RecurrenceTable[{n*(n+4)*a[n] == (5*n^2+14*n+21)*a[n-1] + (5*n^2-4*n+12)*a[n-2] - (n-3)*(n+1)*a[n-3], a[1] == 8, a[2] == 48,a[3] == 264}, a, {n,25}]}] (* _Vaclav Kotesovec_, Oct 05 2012 *)
%Y Fourth diagonal of A033877.
%K nonn
%O 0,2
%A _N. J. A. Sloane_
%E More terms from _Vincenzo Librandi_, May 03 2013