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%I #81 Dec 28 2022 02:22:26
%S 1,3,19,141,1107,8953,73789,616227,5196627,44152809,377379369,
%T 3241135527,27948336381,241813226151,2098240353907,18252025766941,
%U 159114492071763,1389754816243449,12159131877715993,106542797484006471
%N Sum of the squares of the trinomial coefficients (A027907).
%C 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
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 77. (In the integral formula a left bracket is missing for the cosine argument.)
%H Vincenzo Librandi, <a href="/A082758/b082758.txt">Table of n, a(n) for n = 0..200</a>
%H Jelena Đokić, Olga Bodroža-Pantić, and Ksenija Doroslovački, <a href="https://toc.ui.ac.ir/article_27132_9a8d8a500b7f86816f7197d109e4d710.pdf">A spanning union of cycles in rectangular grid graphs, thick grid cylinders and Moebius strips</a>, Transactions on Combinatorics (2023) Art. 27132.
%H Veronika Irvine, <a href="http://hdl.handle.net/1828/7495">Lace Tessellations: A mathematical model for bobbin lace and an exhaustive combinatorial search for patterns</a>, PhD Dissertation, University of Victoria, 2016.
%H Veronika Irvine, Stephen Melczer, and Frank Ruskey, <a href="https://arxiv.org/abs/1804.08725">Vertically constrained Motzkin-like paths inspired by bobbin lace</a>, arXiv:1804.08725 [math.CO], 2018.
%H StackExchange, <a href="http://math.stackexchange.com/questions/1334527/the-asymptotic-behavior-of-an-integral">The asymptotic behavior of an integral</a>, 2015
%F a(n) = Sum_{k=0..2n} T(n, k)^2, where T(n, k) are trinomial coefficients (A027907).
%F a(n) = Sum_{k=0..n} binomial(2*n-k, k)*binomial(2*n, k). - _Benoit Cloitre_, Jul 30 2003
%F 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
%F 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
%F a(n) = Sum_{k=0..n} binomial(n,k)^2*binomial(2*n,n)/ binomial(2*k,k). - _Paul D. Hanna_, Sep 29 2012
%F 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
%F a(n) ~ 3^(2*n+1/2)/(2*sqrt(2*Pi*n)). - _Vaclav Kotesovec_, Oct 14 2012
%F a(n) = GegenbauerC(2*n, -2*n, -1/2). - _Peter Luschny_, May 07 2016
%F From _Peter Luschny_, May 15 2016: (Start)
%F 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.
%F a(n) = hypergeom([1/2-n, -n], [1], 4). (End)
%F a(n) = A002426(2*n). - _Michael Somos_, Jan 08 2017
%F From _Peter Bala_, Mar 16 2018: (Start)
%F a(n) = sqrt(-3)^(2*n)*P(2*n,-1/sqrt(-3)), where P(n,x) is the Legendre polynomia1 of degree n.
%F 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)
%F From _Wolfdieter Lang_, Apr 19 2018 : (Start)
%F 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.
%F 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)
%F From _Peter Bala_, Apr 03 2022: (Start)
%F Conjecture: a(n) = [x^n] ( (1 + x + x^3 + x^4)/(1 - x)^2 )^n.
%F 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. Calculation suggests that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for primes p >= 5 and positive integers n and k.
%F Column 1 of A337389. (End)
%e G.f. = 1 + 3*x + 19*x^2 + 141*x^3 + 1107*x^4 + 8953*x^5 + 73789*x^6 + ...
%p a := n -> simplify(GegenbauerC(2*n,-2*n,-1/2)):
%p seq(a(n), n=0..19); # _Peter Luschny_, May 07 2016
%t 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 *)
%t Table[Hypergeometric2F1[1/2-n,-n,1,4], {n,0,19}] (* _Peter Luschny_, May 15 2016 *)
%t a[ n_] := SeriesCoefficient[ (1 - 2 x - 3 x^2)^(-1/2), {x, 0, 2 n}]; (* _Michael Somos_, Jan 08 2017 *)
%o (PARI) a(n)={local(v=Vec((1+x+x^2)^n));sum(k=1,#v,v[k]^2);}
%o (PARI) a(n)=sum(k=0,n,binomial(2*n-k,k)*binomial(2*n,k));
%o (Maxima) makelist(sum(binomial(2*n-k, k)*binomial(2*n, k),k,0,n),n,0,40);
%o (PARI) {a(n)=sum(k=0,n,binomial(n,k)^2*binomial(2*n,n)/binomial(2*k,k))} \\ _Paul D. Hanna_, Sep 29 2012
%o (Sage)
%o def A():
%o a, b, n = 1, 3, 1
%o yield a
%o while True:
%o yield b
%o n += 1
%o 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))
%o A082758 = A()
%o print([next(A082758) for _ in range(20)]) # _Peter Luschny_, May 16 2016
%Y Cf. A002426, A027907, A273055, A337389.
%K nonn,easy
%O 0,2
%A _Emanuele Munarini_, May 21 2003
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