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Number of lattice paths from (0,0) to (2n-1,n) using steps E=(1,0), N=(0,1), and D=(1,1) which stay weakly above the line through (0,0) and (2n-1,n).
3

%I #25 Jun 23 2023 15:55:34

%S 1,3,17,119,929,7755,67745,611567,5660033,53415251,512072241,

%T 4972855783,48817414177,483649249179,4829637141825,48559914920927,

%U 491195889610241,4995080271452067,51037379418765905,523695644006188887,5394266374440159649,55756104288043890667

%N Number of lattice paths from (0,0) to (2n-1,n) using steps E=(1,0), N=(0,1), and D=(1,1) which stay weakly above the line through (0,0) and (2n-1,n).

%C These are the small nu-Schröder numbers with nu=NE(NEE)^(n-1).

%H M. von Bell and M. Yip, <a href="https://arxiv.org/abs/2006.09804">Schröder combinatorics and nu-associahedra</a>, arXiv:2006.09804 [math.CO], 2020.

%F a(n) = Sum_{i>=0} (1/n)*binomial(2*n-2,i)*binomial(3*n-2-i,2*n-1).

%F a(n) = A108424(n)/2.

%F a(n) ~ phi^(5*n - 1) / (4 * 5^(1/4) * sqrt(Pi) * n^(3/2)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, May 23 2021

%F a(n) = binomial(3*n - 2, 2*n - 1)*hypergeom([2 - 2*n, 1 - n], [2 - 3*n], -1) / n. - _Peter Luschny_, Jun 14 2021

%F D-finite with recurrence (n+1)*(2*n+1)*a(n) +3*(-6*n^2-2*n+1)*a(n-1) +(-46*n^2+135*n-98)*a(n-2) -2*(n-2)*(2*n-5)*a(n-3)=0. - _R. J. Mathar_, Jul 27 2022

%F P-recursive: n*(2*n - 1)*(5*n - 8)*a(n) = (110*n^3 - 396*n^2 + 445*n - 150)*a(n-1) + (n - 2)*(2*n - 5)*(5*n - 3)*a(n-2) with a(1) = 1 and a(2) = 3. - _Peter Bala_, Jun 17 2023

%e For n=2 the a(2)=3 paths are NENE, NDE, and NNEE.

%e For n=3 the a(3)=17 paths are NENEENEE, NENEDEE, NENENEEE, NENDEEE, NENNEEEE, NDEENEE, NDEDEE, NDENEEE, NDDEEE, NDNEEEE, NNEEENEE, NNEEDEE, NNEENEEE, NNEDEEE, NNENEEEE, NNDEEEE, NNNEEEEE.

%p a := n -> binomial(3*n - 2, 2*n - 1)*hypergeom([2 - 2*n, 1 - n], [2 - 3*n], -1)/n:

%p seq(simplify(a(n)), n = 1..22); # _Peter Luschny_, Jun 14 2021

%t Table[Sum[Binomial[2*n - 2, i]*Binomial[3*n - 2 - i, 2*n - 1], {i, 0, 2*n - 2}]/n, {n, 1, 20}] (* _Vaclav Kotesovec_, May 23 2021 *)

%o (PARI) a(n) = {sum(i=0, n, binomial(2*n-2,i)*binomial(3*n-2-i,2*n-1))/n} \\ _Andrew Howroyd_, May 23 2021

%Y Cf. A001622, A108424.

%K nonn,easy

%O 1,2

%A _Matias von Bell_, May 22 2021