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%I #20 Apr 13 2023 09:25:39
%S 1,2,3,6,15,43,133,430,1431,4863,16797,58787,208013,742901,2674441,
%T 9694846,35357671,129644791,477638701,1767263191,6564120421,
%U 24466267021,91482563641,343059613651,1289904147325,4861946401453,18367353072153,69533550916005
%N a(n) = C(n) + 1 - 0^n where C(n) = A000108(n).
%C Hankel transform is A141352.
%C For n >= 2, a(n) is the number of parking functions of size n avoiding the patterns 132, 213, 231, and 312. - _Lara Pudwell_, Apr 12 2023
%H Ayomikun Adeniran and Lara Pudwell, <a href="https://doi.org/10.54550/ECA2023V3S3R17">Pattern avoidance in parking functions</a>, Enumer. Comb. Appl. 3:3 (2023), Article S2R17.
%F G.f.: c(x) + x/(1-x), where c(x) is the g.f. of A000108.
%F Conjecture: (n+1)*a(n) +2*(-3*n+1)*a(n-1) +(9*n-13)*a(n-2) +2*(-2*n+5)*a(n-3)=0. - _R. J. Mathar_, Oct 15 2014
%F a(n) = A000108(n) + A057427(n). - _Alois P. Heinz_, Apr 13 2023
%p a:= n-> signum(n)+binomial(n+n,n)/(n+1):
%p seq(a(n), n=0..30); # _Alois P. Heinz_, Apr 13 2023
%Y Cf. A000108, A057427, A141353.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Jun 27 2008
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