%I #46 Sep 12 2024 11:08:20
%S 1,1,5,32,233,1833,15180,130392,1151057,10378883,95182445,885053524,
%T 8324942620,79071217228,757310811912,7305728683824,70923966744609,
%U 692370887676567,6792525607165935,66933512163735000,662190712902022017,6574831459429388169,65494637699437417584
%N Number of Hyposylvester classes of 3-multiparking functions of length n.
%H J.-C. Novelli and J.-Y. Thibon, <a href="http://arxiv.org/abs/1403.5962">Hopf Algebras of m-permutations, (m+1)-ary trees, and m-parking functions</a>, arXiv preprint arXiv:1403.5962 [math.CO], 2014-2020. See Fig. 27.
%H Jun Yan, <a href="http://arxiv.org/abs/2404.07958">Results on pattern avoidance in parking functions</a>, arXiv preprint arXiv:2404.07958 [math.CO], 2024. See Theorem 4.1.
%F From _Seiichi Manyama_, Aug 12 2023: (Start)
%F The following statements are equivalent:
%F The g.f. satisfies A(x) = 1 + x*A(x)^3 / (1 - 2*x*A(x)^2).
%F a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * binomial(n, k) * binomial(2*n+k+1, n) / (2*n + k + 1).
%F a(n) = (1/n) * Sum_{k=1..n} 3^(n-k) * binomial(n, k) * binomial(2*n, k-1) for n > 0.
%F a(n) = (1/n) * Sum_{k=0..n-1} 2^k * binomial(n, k)*binomial(3*n-k, n-1-k) for n > 0.
%F (End)
%F The above formula is proved in Theorem 4.1 of the Jun Yan link to be the number of Hyposylvester classes of 3-multiparking functions of length n. - _Jun Yan_, Apr 12 2024
%F a(n) ~ 2^(5*n+1) / (sqrt(5*Pi) * n^(3/2) * 3^(n+1)). - _Vaclav Kotesovec_, Apr 12 2024
%F a(n) = 3^(n - 1) * hypergeom([1 - n, -2*n], [2], 1/3) for n > 0. - _Peter Luschny_, Apr 12 2024
%F G.f. A(x) = 1 + series_reversion( x/((1 + 3*x)*(1 + x)^2) ). - _Peter Bala_, Sep 10 2024
%p a := proc(n) option remember; if n <= 1 then return 1 fi;
%p (a(n - 2)*(-800*n^3 + 3024*n^2 - 3184*n + 672) + a(n - 1)*(3275*n^3 - 7467*n^2 +
%p 5038*n - 1008))/(300*n^3 - 234*n^2 - 192*n) end:
%p seq(a(n), n = 0..22); # _Peter Luschny_, Apr 13 2024
%t a[n_] := 3^(n - Boole[n>0]) Hypergeometric2F1[1 - n, -2 n, 2, 1/3];
%t Table[a[n], {n, 0, 22}] (* _Peter Luschny_, Apr 12 2024 *)
%o (PARI) a(n) = sum(k=0, n, 3^k*(-2)^(n-k)*binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1)); \\ _Seiichi Manyama_, Aug 12 2023
%Y Cf. A243694, A243695.
%Y Cf. A007564, A003168, A364923, A364924.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_, Jun 14 2014
%E Name clarified by _Jun Yan_, Apr 12 2024