A084608 -id:A084608 - cp's OEIS frontend

A006139 na(n) = 2(2n-1)a(n-1) + 4(n-1)a(n-2) with a(0) = 1.

1, 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152, 11258054144, 52767312896, 247736643584, 1164829376512, 5484233814016, 25852072517632, 121997903495168

Offset: 0

N. J. A. Sloane

a(n) = number of Delannoy paths (A001850) from (0,0) to (n,n) in which every Northeast step is immediately preceded by an East step. - David Callan, Mar 14 2004
The Hankel transform (see A001906 for definition) of this sequence is A036442 : 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005
In general, 1/sqrt(1-4*r*x-4*r*x^2) has e.g.f. exp(2rx)BesselI(0,2r*sqrt((r+1)/r)x), a(n) = Sum_{k=0..n} C(2k,k)*C(k,n-k)*r^k, gives the central coefficient of (1+(2r)x+r(r+1)x^2) and is the (2r)-th binomial transform of 1/sqrt(1-8*C(n+1,2)x^2). - Paul Barry, Apr 28 2005
Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the H and U steps can have two colors. - N-E. Fahssi, Feb 05 2008
Self-convolution of a(n)/2^n gives Pell numbers A000129(n+1). - Vladimir Reshetnikov, Oct 10 2016
This sequence gives the integer part of an integral approximation to Pi, and also appears in Frits Beukers's "A Rational Approach to Pi" (cf. Links, Example). Despite quality M ~ 0.9058... reported by Beukers, measurements between n = 10000 and 30000 lead to a contentious quality estimate, M ~ 0.79..., at the 99% confidence level. In "Searching for Apéry-Style Miracles" Doron Zeilberger Quotes that M = 0.79119792... and also gives a closed form. The same rational approximation to Pi also follows from time integration on a quartic Hamiltonian surface, 2*H=(q^2+p^2)*(1-4*q*(q-p)). - Bradley Klee, Jul 19 2018, updated Mar 17 2019
Diagonal of rational function 1/(1 - (x + y + x*y^2)). - Gheorghe Coserea, Aug 06 2018

			G.f. = 1 + 2*x + 8*x^2 + 32*x^3 + 136*x^4 + 592*x^5 + 2624*x^6 + 11776*x^7 + ...
J_3 = Integral_{y=0..Pi/4} 4*(4*(sin(y)-cos(y))*sin(y))^3*dy = 32*Pi - (304/3), |J_3| < 1. - _Bradley Klee_, Jul 19 2018

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Frits Beukers, A rational approach to Pi, Nieuw archief voor wiskunde 5/1 No. 4, December 2000, p. 377.
Dario Castellanos, A generalization of Binet's formula and some of its consequences, Fib. Quart., 27 (1989), 424-438.
Maciej Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv:1410.5747 [math.CO], 2014.
Shalosh B. Ekhad and Doron Zeilberger, Searching for Apéry-Style Miracles [Using, Inter-Alia, the Amazing Almkvist-Zeilberger Algorithm], arXiv:1405.4445 [math.NT], 2014.
Bradley Klee, Approximating Pi with Trigonometric-Polynomial Integrals, Wolfram Demonstrations, July 27, 2018.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Cf. A001850, A002426, A036442, A084600-A084606, A084608-A084615.
Cf. A106258, A106259, A106260, A106261.
First column of A110446. A higher-quality Pi approximation: A123178.

a:=[1,2];; for n in [3..25] do a[n]:=1/(n-1)*(2*(2*n-3)*a[n-1]+4*(n-2)*a[n-2]); od; a; # Muniru A Asiru, Aug 06 2018

seq(add(binomial(2*k, k)*binomial(k, n-k), k=0..n), n=0..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 08 2001
A006139 := n -> 2^n*hypergeom([-n/2, 1/2-n/2], [1], 2):
seq(simplify(A006139(n)), n=0..29); # Peter Luschny, Sep 18 2014

Table[SeriesCoefficient[1/(1-4x-4x^2)^(1/2),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 05 2012 *)
Table[Abs[LegendreP[n, I]] 2^n, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 22 2015 *)
Table[Sum[Binomial[2*k, k]*Binomial[k, n - k], {k,0,n}], {n,0,50}] (* G. C. Greubel, Feb 28 2017 *)
a[n_] := If[n == 0, 1, Coefficient[(1 + 2 x + 2 x^2)^n, x^n]] (* Emanuele Munarini, Aug 04 2017 *)
CoefficientList[Series[1/Sqrt[(-4 x^2 - 4 x + 1)], {x, 0, 24}], x] (* Robert G. Wilson v, Jul 28 2018 *)

a(n) := coeff(expand((1+2*x+2*x^2)^n),x,n);
makelist(a(n),n,0,12); /* Emanuele Munarini, Aug 04 2017 */

for(n=0,30,t=polcoeff((1+2*x+2*x^2)^n,n,x); print1(t","))

for(n=0,25, print1(sum(k=0,n, binomial(2*k,k)*binomial(k,n-k)), ", ")) \\ G. C. Greubel, Feb 28 2017

{a(n) = (-2*I)^n * pollegendre(n, I)}; /* Michael Somos, Aug 04 2018 */

a(n) = Sum_{k=0..n} C(2*k, k)*C(k, n-k). - Detlef Pauly (dettodet(AT)yahoo.de), Nov 08 2001
G.f.: 1/(1-4x-4x^2)^(1/2); also, a(n) is the central coefficient of (1+2x+2x^2)^n. - Paul D. Hanna, Jun 01 2003
Inverse binomial transform of central Delannoy numbers A001850. - David Callan, Mar 14 2004
E.g.f.: exp(2*x) * BesselI(0, 2*sqrt(2)*x). - Vladeta Jovovic, Mar 21 2004
a(n) = Sum_{k=0..floor(n/2)} C(n,2k) * C(2k,k) * 2^(n-k). - Paul Barry, Sep 19 2006
a(n) ~ 2^(n - 3/4) * (1 + sqrt(2))^(n + 1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Oct 05 2012, simplified Jan 31 2023
G.f.: 1/(1 - 2*x*(1+x)*Q(0)), where Q(k)= 1 + (4*k+1)*x*(1+x)/(k+1 - x*(1+x)*(2*k+2)*(4*k+3)/(2*x*(1+x)*(4*k+3)+(2*k+3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
a(n) = 2^n*hypergeom([-n/2, 1/2-n/2], [1], 2). - Peter Luschny, Sep 18 2014
0 = a(n)*(+16*a(n+1) + 24*a(n+2) - 8*a(n+3)) + a(n+1)*(+8*a(n+1) + 16*a(n+2) - 6*a(n+3)) + a(n+2)*(-2*a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Oct 13 2016
It appears that Pi/2 = Sum_{n >= 1} (-1)^(n-1)*4^n/(n*a(n-1)*a(n)). - Peter Bala, Feb 20 2017
G.f.: G(x) = (1/(2*Pi))*Integral_{y=0..2*Pi} 1/(1-x*(4*(sin(y)-cos(y))*sin(y)))*dy, also satisfies: (2+4*x)*G(x)-(1-4*x-4*x^2)*G'(x)=0. - Bradley Klee, Jul 19 2018
a(n) = Sum_{k=0..n} (1-i)^k * (1+i)^(n-k) * binomial(n,k)^2, where i is the imaginary unit. - Seiichi Manyama, Aug 29 2025

A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.

Original entry on oeis.org

1, 19, 341, 6119, 109801, 1970299, 35355581, 634430159, 11384387281, 204284540899, 3665737348901, 65778987739319, 1180356041958841, 21180629767519819, 380070979773397901, 6820097006153642399, 122381675130992165281, 2196050055351705332659, 39406519321199703822581

Offset: 0

Clark Kimberling

x(n) := 2*a(n) and y(n) := A007805(n), n >= 0, give all the positive solutions of the Pell equation x^2 - 5*y^2 = -1.
The Gregory V. Richardson formula follows from this. - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
2*a(n) = 2 o 2 o ... o 2 (2*n+1 terms). For example, 2 o 2 = 4*sqrt(5) and 2 o 2 o 2 = 2 o 4*sqrt(5) = 38 = 2*a(1). Cf. A084068.
a(n) = U(2*n+1) where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = sqrt(20)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/4)*( (sqrt(5) + 2)^n - (sqrt(5) - 2)^n ). (End)

			Pell, n=1: (2*19)^2 - 5*17^2 = -1.

G. C. Greubel, Table of n, a(n) for n = 0..795
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Tanya Khovanova, Recursive Sequences.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Eric Weisstein's World of Mathematics, Lehmer Number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials..
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Bisection of A001077 divided by 2.
Cf. A000045, A007805, A049310, A049660, A100047, A188378.
Cf. A005013, A033890, A049685, A084608.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.

[(Fibonacci(6*n+5) - Fibonacci(6*n+1))/4: n in [0..30]]; // G. C. Greubel, Dec 15 2017

with(numtheory): with(combinat):
seq((fibonacci(6*n+5)-fibonacci(6*n+1))/4,n=0..20); # Muniru A Asiru, Mar 25 2018

a[n_] := Simplify[(2 + Sqrt@5)^(2 n - 1) + (2 - Sqrt@5)^(2 n - 1)]/4; Array[a, 16] (* Robert G. Wilson v, Oct 28 2010 *)

my(x='x+O('x^30)); Vec((1+x)/(1 - 18*x + x^2)) \\ G. C. Greubel, Dec 15 2017

a(n) ~ (1/4)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all members x of the sequence, 20*x^2 + 5 is a square. Lim_{n -> oo} a(n)/a(n-1) = 9 + 2*sqrt(20) = 9 + 4*sqrt(5). The 20 can be seen to derive from the statement "20*x^2 + 5 is a square". - Gregory V. Richardson, Oct 12 2002
a(n) = (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)). - Gregory V. Richardson, Oct 12 2002
From R. J. Mathar, Nov 04 2008: (Start)
G.f.: (1+x)/(1 - 18x + x^2).
a(n) = A049660(n) + A049660(n+1). (End)
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=1, a(1)=19. - Philippe Deléham, Nov 17 2008
a(n) = S(n,18) + S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(6*n + 6 - 2*k) + Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) - Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer, k != 1.
The aerated sequence (b(n))n>=1 = [1, 0, 19, 0, 341, 0, 6119, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -16, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = (A188378(n)^3 + (A188378(n)-2)^3) / 8. - Altug Alkan, Jan 24 2016
a(n) = sqrt(5 * Fibonacci(3 + 6*n)^2 - 4)/4. - Gerry Martens, Jul 25 2016
a(n) = Lucas(6*n + 3)/4. - Ehren Metcalfe, Feb 18 2017
From Peter Bala, Mar 23 2018: (Start)
a(n) = 1/4*( (sqrt(5) + 2)^(2*n+1) - (sqrt(5) - 2)^(2*n+1) ).
a(n) = 9*a(n-1) + 2*sqrt(5 + 20*a(n-1)^2).
a(n) = (1/2)*sinh((2*n + 1)*arcsinh(2)). (End)
From Peter Bala, May 09 2025: (Start)
a(n)^2 - 18*a(n)*a(n+1) + a(n+1)^2 = 20.
More generally, for real x, a(n+x)^2 - 18*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 20 with a(n) := (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)) as given above.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/20 (telescoping series).
Product_{n >= 1} ((a(n) + 1)/(a(n) - 1)) = sqrt(5)/2 (telescoping product). (End)

A084609 Coefficients of 1/sqrt(1-4x-8x^2); also, a(n) is the central coefficient of (1+2x+3x^2)^n.

Original entry on oeis.org

1, 2, 10, 44, 214, 1052, 5284, 26840, 137638, 710828, 3692140, 19266920, 100932220, 530479640, 2795917960, 14771797424, 78210099718, 414862155980, 2204273582236, 11729283976136, 62496686731924, 333400654676168

Offset: 0

Paul D. Hanna, Jun 01 2003

Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), U can have 3 colors and H can have 2 colors. - N-E. Fahssi, Mar 30 2008
Self-convolution of a(n)/2^n gives A002605(n+1). - Vladimir Reshetnikov, Oct 10 2016
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 07 2022

Seiichi Manyama, Table of n, a(n) for n = 0..1358 (terms 0..200 from Vincenzo Librandi)
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Cf. A002426, A084600 - A084608, A084610 - A084615.

Row sums of A328347.

A084609:= func< n | (&+[Binomial(n,j)*Binomial(2*(n-j),n)*2^j: j in [0..Floor(n/2)]]) >;
[A084609(n): n in [0..50]]; // G. C. Greubel, Mar 26 2023

(* Programs from Robert G. Wilson v, Mar 02 2011 *)
a[n_]:= Sum[Binomial[n, k] Binomial[2(n-k), n] 2^k, {k, 0, n/2}]; Array[a, 30, 0]
a[n_]:= CoefficientList[Expand[(1 +2x +3x^2)^n], x][[n+1]]; Array[a, 30, 0]
CoefficientList[Series[1/Sqrt[1 -4x -8x^2], {x,0,30}], x]
Range[0, 30]! CoefficientList[ Series[ Exp[ 2x] BesselI[0, Sqrt[12] x], {x, 0, 30}], x] (* End *)
Table[2^n Hypergeometric2F1[(1-n)/2, -n/2, 1, 3], {n,0,30}] (* Vladimir Reshetnikov, Oct 10 2016 *)

a(n):=coeff(expand((1+2*x+3*x^2)^n),x,n);
makelist(a(n),n,0,12);

for(n=0,30,t=polcoeff((1+2*x+3*x^2)^n,n,x); print1(t","))

def A084609(n): return sum(binomial(n,j)*binomial(2*(n-j),n)*2^j for j in range(n//2+1))
[A084609(n) for n in range(51)] # G. C. Greubel, Mar 26 2023

a(n) = Sum_{k = 0..floor(n/2)} C(n,k)*C(2*(n-k),n)*2^k. - Paul Barry, Sep 08 2004
a(n) = Sum_{k = 0..floor(n/2)} C(n,2*k)*C(2*k,k)*3^k*2^(n-2*k); a(n) = Sum_{k = 0..floor(n/2)} C(n,k)*C(n-k,k)*3^k*2^(n-2k). - Paul Barry, Sep 19 2006
E.g.f.: exp(2*x) * Bessel_I(0,2*sqrt(3)*x)
a(n) = ( 2*(2*n-1)*a(n-1) + 8*(n-1)*a(n-2) )/n, a(0)=1, a(1)=2. - Sergei N. Gladkovskii, Jul 20 2012
a(n) ~ sqrt(18+6*sqrt(3))*(2+2*sqrt(3))^n/(6*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
G.f.: 1/(1 - 2*x*(1+2*x)*Q(0)), where Q(k)= 1 + (4*k+1)*x*(1+2*x)/(k+1 - x*(1+2*x)*(2*k+2)*(4*k+3)/(2*x*(1+2*x)*(4*k+3)+(2*k+3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 15 2013
G.f.: G(0), where G(k)= 1 + x*(2+4*x)*(4*k+1)/(2*k+1 - x*(1+2*x)*(2*k+1)*(4*k+3)/(x*(1+2*x)*(4*k+3) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 18 2013
a(n) = 2^n * hypergeom([(1-n)/2,-n/2], [1], 3) = binomial(2*n, n) * hypergeom([(1-n)/2,-n/2], [1/2-n], -2). - Vladimir Reshetnikov, Oct 10 2016
a(n) = (-2*sqrt(-2))^n * P(n, sqrt(-1/2)), where P(n,x) denotes the n-th Legendre polynomial. - Peter Bala, Feb 07 2022

A084606 Triangle, read by rows, where the n-th row lists the (2n+1) coefficients of (1+2x+2x^2)^n.

Original entry on oeis.org

1, 1, 2, 2, 1, 4, 8, 8, 4, 1, 6, 18, 32, 36, 24, 8, 1, 8, 32, 80, 136, 160, 128, 64, 16, 1, 10, 50, 160, 360, 592, 720, 640, 400, 160, 32, 1, 12, 72, 280, 780, 1632, 2624, 3264, 3120, 2240, 1152, 384, 64, 1, 14, 98, 448, 1484, 3752, 7448, 11776, 14896, 15008, 11872

Offset: 0

Paul D. Hanna, Jun 01 2003

			Rows:
{1},
{1,2,2},
{1,4,8,8,4},
{1,6,18,32,36,24,8},
{1,8,32,80,136,160,128,64,16},
{1,10,50,160,360,592,720,640,400,160,32},
{1,12,72,280,780,1632,2624,3264,3120,2240,1152,384,64},

Cf. A002426, A084600-A084605, A006139, A084608-A084615.

for(n=0,15, for(k=0,2*n,t=polcoeff((1+2*x+2*x^2)^n,k,x); print1(t",")); print(" "))

A329708 Triangle, read by rows, where the n-th row lists the (2n+1) coefficients of (1+2x+...+(n+1)x^n)^2.

Original entry on oeis.org

1, 1, 4, 4, 1, 4, 10, 12, 9, 1, 4, 10, 20, 25, 24, 16, 1, 4, 10, 20, 35, 44, 46, 40, 25, 1, 4, 10, 20, 35, 56, 70, 76, 73, 60, 36, 1, 4, 10, 20, 35, 56, 84, 104, 115, 116, 106, 84, 49, 1, 4, 10, 20, 35, 56, 84, 120, 147, 164, 170, 164, 145, 112, 64

Offset: 0

Seiichi Manyama, Feb 29 2020

			Triangle begins:
  1;
  1, 4,  4;
  1, 4, 10, 12,  9;
  1, 4, 10, 20, 25, 24, 16;
  1, 4, 10, 20, 35, 44, 46, 40, 25;
  ...

Seiichi Manyama, Rows n = 0..99, flattened

Row sums give A000537(n+1).
T(n,2n) gives A000290(n+1).
Cf. A000292, A002260, A084608, A008794, A329709.

row[n_]:=CoefficientList[Series[(Sum[(i+1)x^i,{i,0,n}])^2,{x,0,2n}],x]; Array[row,8,0]//Flatten (* Stefano Spezia, Feb 15 2025 *)

for(n=0, 10, print(Vecrev(sum(k=0, n, (k+1)*x^k)^2), ", "))

T(n,k) = A000292(k+1) for k=0..n.

Sum_{k=0..2n} (-1)^k * T(n,k) = A008794(n+2). - Alois P. Heinz, Feb 14 2025

cp's OEIS Frontend

A006139 n*a(n) = 2*(2*n-1)*a(n-1) + 4*(n-1)*a(n-2) with a(0) = 1.

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A049629 a(n) = (F(6*n+5) - F(6*n+1))/4 = (F(6*n+4) + F(6*n+2))/4, where F = A000045.

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A084609 Coefficients of 1/sqrt(1-4*x-8*x^2); also, a(n) is the central coefficient of (1+2*x+3*x^2)^n.

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A084606 Triangle, read by rows, where the n-th row lists the (2n+1) coefficients of (1+2x+2x^2)^n.

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A329708 Triangle, read by rows, where the n-th row lists the (2n+1) coefficients of (1+2*x+...+(n+1)*x^n)^2.

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A006139 na(n) = 2(2n-1)a(n-1) + 4(n-1)a(n-2) with a(0) = 1.

A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.

A084609 Coefficients of 1/sqrt(1-4x-8x^2); also, a(n) is the central coefficient of (1+2x+3x^2)^n.

A329708 Triangle, read by rows, where the n-th row lists the (2n+1) coefficients of (1+2x+...+(n+1)x^n)^2.