A273055 -id:A273055 - cp's OEIS frontend

A002426 Central trinomial coefficients: largest coefficient of (1 + x + x^2)^n.

1, 1, 3, 7, 19, 51, 141, 393, 1107, 3139, 8953, 25653, 73789, 212941, 616227, 1787607, 5196627, 15134931, 44152809, 128996853, 377379369, 1105350729, 3241135527, 9513228123, 27948336381, 82176836301, 241813226151, 712070156203, 2098240353907, 6186675630819

Offset: 0

N. J. A. Sloane, Simon Plouffe

Number of ordered trees with n + 1 edges, having root of odd degree and nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 02 2002
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), running from (0,0) to (n,0) (i.e., grand Motzkin paths of length n). For example, a(3) = 7 because we have HHH, HUD, HDU, UDH, DUH, UHD and DHU. - Emeric Deutsch, May 31 2003
Number of lattice paths from (0,0) to (n,n) using steps (2,0), (0,2), (1,1). It appears that 1/sqrt((1 - x)^2 - 4*x^s) is the g.f. for lattice paths from (0,0) to (n,n) using steps (s,0), (0,s), (1,1). - Joerg Arndt, Jul 01 2011
Number of lattice paths from (0,0) to (n,n) using steps (1,0), (1,1), (1,2). - Joerg Arndt, Jul 05 2011
Binomial transform of A000984, with interpolated zeros. - Paul Barry, Jul 01 2003
Number of leaves in all 0-1-2 trees with n edges, n > 0. (A 0-1-2 tree is an ordered tree in which every vertex has at most two children.) - Emeric Deutsch, Nov 30 2003
a(n) is the number of UDU-free paths of n + 1 upsteps (U) and n downsteps (D) that start U. For example, a(2) = 3 counts UUUDD, UUDDU, UDDUU. - David Callan, Aug 18 2004
Diagonal sums of triangle A063007. - Paul Barry, Aug 31 2004
Number of ordered ballots from n voters that result in an equal number of votes for candidates A and B in a three candidate election. Ties are counted even when candidates A and B lose the election. For example, a(3) = 7 because ballots of the form (voter-1 choice, voter-2 choice, voter-3 choice) that result in equal votes for candidates A and B are the following: (A,B,C), (A,C,B), (B,A,C), (B,C,A), (C,A,B), (C,B,A) and (C,C,C). - Dennis P. Walsh, Oct 08 2004
a(n) is the number of weakly increasing sequences (a_1,a_2,...,a_n) with each a_i in [n]={1,2,...,n} and no element of [n] occurring more than twice. For n = 3, the sequences are 112, 113, 122, 123, 133, 223, 233. - David Callan, Oct 24 2004
Note that n divides a(n+1) - a(n). In fact, (a(n+1) - a(n))/n = A007971(n+1). - T. D. Noe, Mar 16 2005
Row sums of triangle A105868. - Paul Barry, Apr 23 2005
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having no H steps on the x-axis. Example: a(3) = 7 because we have UDU, UHD, UHH, UHU, UUD, UUH and UUU. - Emeric Deutsch, Oct 07 2007
Equals right border of triangle A152227; starting with offset 1, the row sums of triangle A152227. - Gary W. Adamson, Nov 29 2008
Starting with offset 1 = iterates of M * [1,1,1,...] where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
a(n) is prime for n = 2, 3 and 4, with no others for n <= 10^5 (E. W. Weisstein, Mar 14 2005). It has apparently not been proved that no [other] prime central trinomials exist. - Jonathan Vos Post, Mar 19 2010
a(n) is not divisible by 3 for n whose base-3 representation contains no 2 (A005836).
a(n) = number of (n-1)-lettered words in the alphabet {1,2,3} with as many occurrences of the substring (consecutive subword) [1,2] as those of [2,1]. See the papers by Ekhad-Zeilberger and Zeilberger. - N. J. A. Sloane, Jul 05 2012
a(n) = coefficient of x^n in (1 + x + x^2)^n. - L. Edson Jeffery, Mar 23 2013
a(n) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that (i.) A and B are disjoint and (ii.) A and B contain the same number of elements. For example, a(2) = 3 because we have: ({},{}) ; ({1},{2}) ; ({2},{1}). - Geoffrey Critzer, Sep 04 2013
Also central terms of A082601. - Reinhard Zumkeller, Apr 13 2014
a(n) is the number of n-tuples with entries 0, 1, or 2 and with the sum of entries equal to n. For n=3, the seven 3-tuples are (1,1,1), (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), and (2,1,0). - Dennis P. Walsh, May 08 2015
The series 2*a(n) + 3*a(n+1) + a(n+2) = 2*A245455(n+3) has Hankel transform of L(2n+1)*2^n, offset n = 1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
The series (2*a(n) + 3*a(n+1) + a(n+2))/2 = A245455(n+3) has Hankel transform of L(2n+1), offset n=1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
Conjecture: An integer n > 3 is prime if and only if a(n) == 1 (mod n^2). We have verified this for n up to 8*10^5, and proved that a(p) == 1 (mod p^2) for any prime p > 3 (cf. A277640). - Zhi-Wei Sun, Nov 30 2016
This is the analog for Coxeter type B of Motzkin numbers (A001006) for Coxeter type A. - F. Chapoton, Jul 19 2017
a(n) is also the number of solutions to the equation x(1) + x(2) + ... + x(n) = 0, where x(1), ..., x(n) are in the set {-1,0,1}. Indeed, the terms in (1 + x + x^2)^n that produce x^n are of the form x^i(1)*x^i(2)*...*x^i(n) where i(1), i(2), ..., i(n) are in {0,1,2} and i(1) + i(2) + ... + i(n) = n. By setting j(t) = i(t) - 1 we obtain that j(1), ..., j(n) satisfy j(1) + ... + j(n) =0 and j(t) in {-1,0,1} for all t = 1..n. - Lucien Haddad, Mar 10 2018
If n is a prime greater than 3 then a(n)-1 is divisible by n^2. - Ira M. Gessel, Aug 08 2021
Let f(m) = ceiling((q+log(q))/log(9)), where q = -log(log(27)/(2*m^2*Pi)) then f(a(n)) = n, for n > 0. - Miko Labalan, Oct 07 2024
Diagonal of the rational function 1 / (1 - x^2 - y^2 - x*y). - Ilya Gutkovskiy, Apr 23 2025

			For n = 2, (x^2 + x + 1)^2 = x^4 + 2*x^3 + 3*x^2 + 2*x + 1, so a(2) = 3. - _Michael B. Porter_, Sep 06 2016

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Index entries for sequences of k-nomial coefficients
Index entries for "core" sequences

INVERT transform is A007971. Partial sums are A097893. Squares are A168597.
Main column of A027907. Column k=2 of A305161. Column k=0 of A328347. Column 1 of A201552(?).
Cf. A001006, A002878, A005043, A005717, A082758 (bisection), A273055 (bisection), A102445, A113302, A113303, A113304, A113305 (divisibility of central trinomial coefficients), A152227, A277640.

a002426 n = a027907 n n  -- Reinhard Zumkeller, Jan 22 2013

P:=PolynomialRing(Integers()); [Max(Coefficients((1+x+x^2)^n)): n in [0..26]]; // Bruno Berselli, Jul 05 2011

A002426 := proc(n) local k;
    sum(binomial(n, k)*binomial(n-k, k), k=0..floor(n/2));
end proc: # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
# Alternatively:
a := n -> simplify(GegenbauerC(n,-n,-1/2)):
seq(a(n), n=0..29); # Peter Luschny, May 07 2016

Table[ CoefficientList[ Series[(1 + x + x^2)^n, {x, 0, n}], x][[ -1]], {n, 0, 27}] (* Robert G. Wilson v *)
a=b=1; Join[{a,b}, Table[c=((2n-1)b + 3(n-1)a)/n; a=b; b=c; c, {n,2,100}]]; Table[Sqrt[-3]^n LegendreP[n,1/Sqrt[-3]],{n,0,26}] (* Wouter Meeussen, Feb 16 2013 *)
a[ n_] := If[ n < 0, 0, 3^n Hypergeometric2F1[ 1/2, -n, 1, 4/3]]; (* Michael Somos, Jul 08 2014 *)
Table[4^n *JacobiP[n,-n-1/2,-n-1/2,-1/2], {n,0,29}] (* Peter Luschny, May 13 2016 *)
a[n_] := a[n] = Sum[n!/((n - 2*i)!*(i!)^2), {i, 0, n/2}]; Table[a[n], {n, 0, 29}] (* Shara Lalo and Zagros Lalo, Oct 03 2018 *)

trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
makelist(trinomial(n,n),n,0,12); /* Emanuele Munarini, Mar 15 2011 */

makelist(ultraspherical(n,-n,-1/2),n,0,12); /* Emanuele Munarini, Dec 20 2016 */

{a(n) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, n))};

/* as lattice paths: same as in A092566 but use */
steps=[[2, 0], [0, 2], [1, 1]];
/* Joerg Arndt, Jul 01 2011 */

a(n)=polcoeff(sum(m=0, n, (2*m)!/m!^2 * x^(2*m) / (1-x+x*O(x^n))^(2*m+1)), n) \\ Paul D. Hanna, Sep 21 2013

from math import comb
def A002426(n): return sum(comb(n,k)*comb(k,n-k) for k in range(n+1)) # Chai Wah Wu, Nov 15 2022

A002426 = lambda n: hypergeometric([-n/2, (1-n)/2], [1], 4)
[simplify(A002426(n)) for n in (0..29)]
# Peter Luschny, Sep 17 2014

def A():
    a, b, n = 1, 1, 1
    yield a
    while True:
        yield b
        n += 1
        a, b = b, ((3 * (n - 1)) * a + (2 * n - 1) * b) // n
A002426 = A()
print([next(A002426) for  in range(30)])  # _Peter Luschny, May 16 2016

G.f.: 1/sqrt(1 - 2*x - 3*x^2).
E.g.f.: exp(x)*I_0(2x), where I_0 is a Bessel function. - Michael Somos, Sep 09 2002
a(n) = 2*A027914(n) - 3^n. - Benoit Cloitre, Sep 28 2002
a(n) is asymptotic to d*3^n/sqrt(n) with d around 0.5.. - Benoit Cloitre, Nov 02 2002, d = sqrt(3/Pi)/2 = 0.4886025119... - Alec Mihailovs (alec(AT)mihailovs.com), Feb 24 2005
D-finite with recurrence: a(n) = ((2*n - 1)*a(n-1) + 3*(n - 1)*a(n-2))/n; a(0) = a(1) = 1; see paper by Barcucci, Pinzani and Sprugnoli.
Inverse binomial transform of A000984. - Vladeta Jovovic, Apr 28 2003
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, k/2)*(1 + (-1)^k)/2; a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*binomial(2*k, k). - Paul Barry, Jul 01 2003
a(n) = Sum_{k>=0} binomial(n, 2*k)*binomial(2*k, k). - Philippe Deléham, Dec 31 2003
a(n) = Sum_{i+j=n, 0<=j<=i<=n} binomial(n, i)*binomial(i, j). - Benoit Cloitre, Jun 06 2004
a(n) = 3*a(n-1) - 2*A005043(n). - Joost Vermeij (joost_vermeij(AT)hotmail.com), Feb 10 2005
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, n-k). - Paul Barry, Apr 23 2005
a(n) = (-1/4)^n*Sum_{k=0..n} binomial(2*k, k)*binomial(2*n-2*k, n-k)*(-3)^k. - Philippe Deléham, Aug 17 2005
a(n) = A111808(n,n). - Reinhard Zumkeller, Aug 17 2005
a(n) = Sum_{k=0..n} (((1 + (-1)^k)/2)*Sum_{i=0..floor((n-k)/2)} binomial(n, i)*binomial(n-i, i+k)*((k + 1)/(i + k + 1))). - Paul Barry, Sep 23 2005
a(n) = 3^n*Sum_{j=0..n} (-1/3)^j*C(n, j)*C(2*j, j); follows from (a) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/2)^n*Sum_{j=0..n} 3^j*binomial(n, j)*binomial(2*n-2*j, n) = (3/2)^n*Sum_{j=0..n} (1/3)^j*binomial(n, j)*binomial(2*j, n); follows from (c) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/Pi)*Integral_{x=-1..3} x^n/sqrt((3 - x)*(1 + x)) is moment representation. - Paul Barry, Sep 10 2007
G.f.: 1/(1 - x - 2x^2/(1 - x - x^2/(1 - x - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 05 2009
a(n) = sqrt(-1/3)*(-1)^n*hypergeometric([1/2, n+1], [1], 4/3). - Mark van Hoeij, Nov 12 2009
a(n) = (1/Pi)*Integral_{x=-1..1} (1 + 2*x)^n/sqrt(1 - x^2) = (1/Pi)*Integral_{t=0..Pi} (1 + 2*cos(t))^n. - Eli Wolfhagen, Feb 01 2011
In general, g.f.: 1/sqrt(1 - 2*a*x + x^2*(a^2 - 4*b)) = 1/(1 - a*x)*(1 - 2*x^2*b/(G(0)*(a*x - 1) + 2*x^2*b)); G(k) = 1 - a*x - x^2*b/G(k+1); for g.f.: 1/sqrt(1 - 2*x - 3*x^2) = 1/(1 - x)*(1 - 2*x^2/(G(0)*(x - 1) + 2*x^2)); G(k) = 1 - x - x^2/G(k+1), a = 1, b = 1; (continued fraction). - Sergei N. Gladkovskii, Dec 08 2011
a(n) = Sum_{k=0..floor(n/3)} (-1)^k*binomial(2*n-3*k-1, n-3*k)*binomial(n, k). - Gopinath A. R., Feb 10 2012
G.f.: A(x) = x*B'(x)/B(x) where B(x) satisfies B(x) = x*(1 + B(x) + B(x)^2). - Vladimir Kruchinin, Feb 03 2013 (B(x) = x*A001006(x) - Michael Somos, Jul 08 2014)
G.f.: G(0), where G(k) = 1 + x*(2 + 3*x)*(4*k + 1)/(4*k + 2 - x*(2 + 3*x)*(4*k + 2)*(4*k + 3)/(x*(2 + 3*x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
E.g.f.: exp(x) * Sum_{k>=0} (x^k/k!)^2. - Geoffrey Critzer, Sep 04 2013
G.f.: Sum_{n>=0} (2*n)!/n!^2*(x^(2*n)/(1 - x)^(2*n+1)). - Paul D. Hanna, Sep 21 2013
0 = a(n)*(9*a(n+1) + 9*a(n+2) - 6*a(n+3)) + a(n+1)*(3*a(n+1) + 4*a(n+2) - 3*a(n+3)) + a(n+2)*(-a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Jul 08 2014
a(n) = hypergeometric([-n/2, (1-n)/2], [1], 4). - Peter Luschny, Sep 17 2014
a(n) = A132885(n,0), that is, a(n) = A132885(A002620(n+1)). - Altug Alkan, Nov 29 2015
a(n) = GegenbauerC(n,-n,-1/2). - Peter Luschny, May 07 2016
a(n) = 4^n*JacobiP[n,-n-1/2,-n-1/2,-1/2]. - Peter Luschny, May 13 2016
From Alexander Burstein, Oct 03 2017: (Start)
G.f.: A(4*x) = B(-x)*B(3*x), where B(x) is the g.f. of A000984.
G.f.: A(2*x)*A(-2*x) = B(x^2)*B(9*x^2).
G.f.: A(x) = 1 + x*M'(x)/M(x), where M(x) is the g.f. of A001006. (End)
a(n) = Sum_{i=0..n/2} n!/((n - 2*i)!*(i!)^2). [Cf. Lalo and Lalo link. It is Luschny's terminating hypergeometric sum.] - Shara Lalo and Zagros Lalo, Oct 03 2018
From Peter Bala, Feb 07 2022: (Start)
a(n)^2 = Sum_{k = 0..n} (-3)^(n-k)*binomial(2*k,k)^2*binomial(n+k,n-k) and has g.f. Sum_{n >= 0} binomial(2*n,n)^2*x^n/(1 + 3*x)^(2*n+1). Compare with the g.f. for a(n) given above by Hanna.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all prime p and positive integers n and k.
Conjecture: The stronger congruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all prime p >= 5 and positive integers n and k. (End)
a(n) = A005043(n) + A005717(n) for n >= 1. - Amiram Eldar, May 17 2024
For even n, a(n) = (n-1)!!* 2^{n/2}/ (n/2)!* 2F1(-n/2,-n/2;1/2;1/4). For odd n, a(n) = n!! *2^(n/2-1/2) / (n/2-1/2)! * 2F1(1/2-n/2,1/2-n/2;3/2;1/4). - R. J. Mathar, Mar 19 2025

A082758 Sum of the squares of the trinomial coefficients (A027907).

Original entry on oeis.org

1, 3, 19, 141, 1107, 8953, 73789, 616227, 5196627, 44152809, 377379369, 3241135527, 27948336381, 241813226151, 2098240353907, 18252025766941, 159114492071763, 1389754816243449, 12159131877715993, 106542797484006471, 934837217271732457, 8212609533895771131

Offset: 0

Emanuele Munarini, May 21 2003

a(n) = T(2*n, 2*n), the coefficient of x^(2*n) in (1+x+x^2)^(2*n), where T is the trinomial triangle A027907; Integral representation: a(n) = (1/Pi) * Integral_{x=-1..1} ((1+2*x)^(2*n)/sqrt(1-x^2)), i.e., a(n) is the moment of order 2n of the random variable 1+2X, where the distribution of X is an arcsin law on the interval (-1,1). - N-E. Fahssi, Jan 22 2008

			G.f. = 1 + 3*x + 19*x^2 + 141*x^3 + 1107*x^4 + 8953*x^5 + 73789*x^6 + ...

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 77. (In the integral formula a left bracket is missing for the cosine argument.)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jelena Đokić, Olga Bodroža-Pantić, and Ksenija Doroslovački, A spanning union of cycles in rectangular grid graphs, thick grid cylinders and Moebius strips, Transactions on Combinatorics (2023) Art. 27132.
Veronika Irvine, Lace Tessellations: A mathematical model for bobbin lace and an exhaustive combinatorial search for patterns, PhD Dissertation, University of Victoria, 2016.
Veronika Irvine, Stephen Melczer, and Frank Ruskey, Vertically constrained Motzkin-like paths inspired by bobbin lace, arXiv:1804.08725 [math.CO], 2018.
StackExchange, The asymptotic behavior of an integral, 2015.

Cf. A002426, A027907, A132303 - A132305, A168592, A273055, A337389.

a := n -> simplify(GegenbauerC(2*n,-2*n,-1/2)):
seq(a(n), n=0..19); # Peter Luschny, May 07 2016

Table[Sum[(-1)^(i)*Binomial[2*n,i]*Binomial[4*n-3*i-1,2*n-3*i],{i,0,2*n/3}],{n, 0,25}] (* Adi Dani, Jul 03 2011 *)
Table[Hypergeometric2F1[1/2-n,-n,1,4], {n,0,19}] (* Peter Luschny, May 15 2016 *)
a[ n_] := SeriesCoefficient[ (1 - 2 x - 3 x^2)^(-1/2), {x, 0, 2 n}]; (* Michael Somos, Jan 08 2017 *)

makelist(sum(binomial(2*n-k, k)*binomial(2*n, k),k,0,n),n,0,40);

a(n)={local(v=Vec((1+x+x^2)^n));sum(k=1,#v,v[k]^2);}

a(n)=sum(k=0,n,binomial(2*n-k,k)*binomial(2*n,k));

{a(n)=sum(k=0,n,binomial(n,k)^2*binomial(2*n,n)/binomial(2*k,k))} \\ Paul D. Hanna, Sep 29 2012

def A():
    a, b, n = 1, 3, 1
    yield a
    while True:
        yield b
        n += 1
        a, b = b, ((9-9*n)*(4*n-1)*(2*n-3)*a+(4*n-3)*(20*n^2-30*n+7)*b)//(n*(2*n-1)*(4*n-5))
A082758 = A()
print([next(A082758) for  in range(20)]) # _Peter Luschny, May 16 2016

a(n) = Sum_{k=0..2n} T(n, k)^2, where T(n, k) are trinomial coefficients (A027907).
a(n) = Sum_{k=0..n} binomial(2*n-k, k)*binomial(2*n, k). - Benoit Cloitre, Jul 30 2003
G.f.: (1/sqrt(1+2*x-3*x^2) + 1/sqrt(1-2*x-3*x^2))/2 (with interpolated zeros). - Paul Barry, Jan 04 2005
a(n) = Sum_{k=0..n} binomial(2*n,2*k)*binomial(2*k,k) = Sum_{k=0..n} binomial(n+k,2k)*binomial(2*n,n+k). - Paul Barry, Dec 16 2008
a(n) = Sum_{k=0..n} binomial(n,k)^2*binomial(2*n,n)/ binomial(2*k,k). - Paul D. Hanna, Sep 29 2012
Recurrence: n*(2*n-1)*a(n) = (14*n^2+n-12)*a(n-1) + 3*(14*n^2-71*n+78)*a(n-2) - 27*(n-2)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ 3^(2*n+1/2)/(2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = GegenbauerC(2*n, -2*n, -1/2). - Peter Luschny, May 07 2016
From Peter Luschny, May 15 2016: (Start)
a(n) = ((9-9*n)*(4*n-1)*(2*n-3)*a(n-2)+(4*n-3)*(20*n^2-30*n+7)*a(n-1))/(n*(2*n-1)*(4*n-5)) for n>=2.
a(n) = hypergeom([1/2-n, -n], [1], 4). (End)
a(n) = A002426(2*n). - Michael Somos, Jan 08 2017
From Peter Bala, Mar 16 2018: (Start)
a(n) = sqrt(-3)^(2*n)*P(2*n,-1/sqrt(-3)), where P(n,x) is the Legendre polynomial of degree n.
a(n) = 1/C(2*n,n)*Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k)*(-3)^(n-k). Cf. A273055. (End)
From Wolfdieter Lang, Apr 19 2018 : (Start)
a(n) = (2/Pi)*Integral_{phi=0..Pi/2} (sin(3*phi)/sin(phi))^(2*n) [Comtet, p. 77, q=3, n=k -> 2*n] = (2/Pi)*Integral_{x=0..2} (x^2 - 1)^(2*n)/sqrt(4-x^2) (with x = 2*cos(phi)). See also the integral of the above comment.
a(n) = 3^(2*n)*Sum_{k=0..2*n} binomial(2*n, k)*binomial(2*k, k)*(-1/3)^k = 3^(2*n)*hypergeometric([-2*n, 1/2], [1], 4/3) = (-3)^n*LegendreP(2*n, 1/sqrt(-3)). (End)
From Peter Bala, Apr 03 2022: (Start)
Conjecture: a(n) = [x^n] ( (1 + x + x^3 + x^4)/(1 - x)^2 )^n.
If the conjecture is true then the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. [added Aug 29 2025: The conjecture is true. The conjecture leads to the double sum representation a(n) = Sum_{j, k} binomial(n, k)*binomial(n, j)*binomial(3*n-3*j-k-1, n-3*j-k), which satisfies the second-order recurrence given above by Peter Luschny, as can be verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger Maple package.]
Calculation suggests that the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) holds for primes p >= 5 and positive integers n and k.
Column 1 of A337389. (End)

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A002426 Central trinomial coefficients: largest coefficient of (1 + x + x^2)^n.

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A082758 Sum of the squares of the trinomial coefficients (A027907).

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A337369 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1-2*(k+4)*x+((k-4)*x)^2) * (1+(k-4)*x+sqrt(1-2*(k+4)*x+((k-4)*x)^2)) )).

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A337394 Expansion of sqrt(2 / ( (1-6*x+25*x^2) * (1-5*x+sqrt(1-6*x+25*x^2)) )).

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A281000 Triangle read by rows: T(n,k) = binomial(2*n+1, 2*k+1)*binomial(2*n-2*k, n-k)/(n+1-k) for 0 <= k <= n.

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A291081 Irregular triangle read by rows: T(n,m) = number of lattice paths of type A^H terminating at point (n, m).

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A337369 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1-2(k+4)x+((k-4)x)^2) (1+(k-4)x+sqrt(1-2(k+4)x+((k-4)x)^2)) )).

A337394 Expansion of sqrt(2 / ( (1-6x+25x^2) * (1-5x+sqrt(1-6x+25*x^2)) )).

A281000 Triangle read by rows: T(n,k) = binomial(2n+1, 2k+1)binomial(2n-2*k, n-k)/(n+1-k) for 0 <= k <= n.