A082758 Sum of the squares of the trinomial coefficients (A027907).
1, 3, 19, 141, 1107, 8953, 73789, 616227, 5196627, 44152809, 377379369, 3241135527, 27948336381, 241813226151, 2098240353907, 18252025766941, 159114492071763, 1389754816243449, 12159131877715993, 106542797484006471, 934837217271732457, 8212609533895771131
Offset: 0
Examples
G.f. = 1 + 3*x + 19*x^2 + 141*x^3 + 1107*x^4 + 8953*x^5 + 73789*x^6 + ...
References
- L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 77. (In the integral formula a left bracket is missing for the cosine argument.)
Links
- 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.
Programs
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Maple
a := n -> simplify(GegenbauerC(2*n,-2*n,-1/2)): seq(a(n), n=0..19); # Peter Luschny, May 07 2016
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Mathematica
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 *) -
Maxima
makelist(sum(binomial(2*n-k, k)*binomial(2*n, k),k,0,n),n,0,40);
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PARI
a(n)={local(v=Vec((1+x+x^2)^n));sum(k=1,#v,v[k]^2);} -
PARI
a(n)=sum(k=0,n,binomial(2*n-k,k)*binomial(2*n,k));
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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 -
Sage
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
Formula
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)
Comments