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a(n) = T(2n, n), where T is the Delannoy triangle (A008288).
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%I #49 Sep 09 2024 21:41:53

%S 1,5,41,377,3649,36365,369305,3800305,39490049,413442773,4354393801,

%T 46082942185,489658242241,5220495115997,55818956905529,

%U 598318746037217,6427269150511105,69175175263888037,745778857519239785,8052432236270744665,87063177396677721409

%N a(n) = T(2n, n), where T is the Delannoy triangle (A008288).

%C Even order terms in the diagonal of rational function 1/(1 - (x + y^2 + x*y^2)). - _Gheorghe Coserea_, Aug 31 2018

%H Vincenzo Librandi, <a href="/A026000/b026000.txt">Table of n, a(n) for n = 0..200</a>

%H Lin Yang, Yu-Yuan Zhang, and Sheng-Liang Yang, <a href="https://doi.org/10.1016/j.laa.2023.12.021">The halves of Delannoy matrix and Chung-Feller properties of the m-Schröder paths</a>, Linear Alg. Appl. (2024).

%F a(n) = ((2*n+3)*(n+1)*A027307(n+1)/2-(3*n+2)*n*A027307(n)) / (5*n+3) (guessed). - _Mark van Hoeij_, Jul 02 2010

%F Recurrence: 2*n*(2*n-1)*a(n) = (46*n^2-51*n+15)*a(n-1) - (18*n^2-82*n+85)*a(n-2) - (n-2)*(2*n-5)*a(n-3). - _Vaclav Kotesovec_, Oct 08 2012

%F a(n) ~ sqrt(150+70*sqrt(5))*((11+5*sqrt(5))/2)^n/(20*sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 08 2012. Equivalently, a(n) ~ phi^(5*n + 2) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Dec 08 2021

%F a(n) = hypergeom([-n, -n, n + 1], [1/2, 1], 1). - _Peter Luschny_, Mar 14 2018

%F From _Gheorghe Coserea_, Aug 31 2018:(Start)

%F G.f.: 1 + serreverse((-(44*x^2 + 88*x + 45) + (10*x + 9)*sqrt(20*x^2 + 44*x + 25))/(8*(x + 1)^2)).

%F G.f. y=A(x) satisfies:

%F 0 = 4*(x^2 + 11*x - 1)*y^3 + (x + 3)*y + 1.

%F 0 = 2*x*(x - 2)*(x^2 + 11*x - 1)*y'' + (5*x^3 + 8*x^2 - 87*x + 2)*y' + (x^2 - 7*x - 10)*y. (End)

%F From _Peter Bala_, Jan 20 2020: (Start)

%F a(n) = Sum_{k = 0..n} C(2*n, n-k) * C(2*n+k, k).

%F a(n) = C(2*n, n) * hypergeom([-n, 2*n+1], [n+1], -1).

%F n*(2*n-1)*(10*n-13)*a(n) = (220*n^3-506*n^2+334*n-63*n)*a(n-1) + (n-1)*(2*n-3)*(10*n-3)*a(n-2). (End)

%F From _Peter Bala_, Apr 15 2023: (Start)

%F a(n) = Sum_{k = 0..n} binomial(n, k)*binomial(2*n, k)*2^k

%F a(n) = (-1)^n * Sum_{k = 0..n} binomial(n, k)*binomial(2*n+k, k)*(-2)^k.

%F a(n) = hypergeom([-n, -2*n], [1], 2) = (-1)^n * hypergeom([-n, 2*n + 1], [1], 2). (End)

%e A(x) = 1 + 5*x + 41*x^2 + 377*x^3 + 3649*x^4 + 36365*x^5 + ...

%t Flatten[{1,RecurrenceTable[{2*n*(2*n-1)*a[n] == (46*n^2-51*n+15)*a[n-1] - (18*n^2-82*n+85)*a[n-2] - (n-2)*(2*n-5)*a[n-3],a[1]==5,a[2]==41,a[3]==377},a,{n,20}]}] (* _Vaclav Kotesovec_, Oct 08 2012 *)

%t a[n_] := HypergeometricPFQ[{-n, -n, n + 1}, {1/2, 1}, 1];

%t Table[a[n], {n, 0, 18}] (* _Peter Luschny_, Mar 14 2018 *)

%o (PARI)

%o seq(N) = {

%o my(a = vector(N)); a[1]=5; a[2]=41; a[3]=377;

%o for (n=4, N,

%o a[n] = (46*n^2-51*n+15)*a[n-1] - (18*n^2-82*n+85)*a[n-2] - (n-2)*(2*n-5)*a[n-3];

%o a[n] /= 2*n*(2*n-1));

%o concat(1, a);

%o };

%o seq(18)

%o \\ test: y=Ser(seq(303),'x); 0 == 4*(x^2 + 11*x - 1)*y^3 + (x + 3)*y + 1

%o \\ _Gheorghe Coserea_, Aug 31 2018

%Y Cf. A008288, A027307, A026001.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_