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A082758
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Sum of the squares of the trinomial coefficients (A027907).
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15
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1, 3, 19, 141, 1107, 8953, 73789, 616227, 5196627, 44152809, 377379369, 3241135527, 27948336381, 241813226151, 2098240353907, 18252025766941, 159114492071763, 1389754816243449, 12159131877715993, 106542797484006471
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OFFSET
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0,2
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COMMENTS
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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
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REFERENCES
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L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 77. (In the integral formula a left bracket is missing for the cosine argument.)
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LINKS
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FORMULA
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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) = GegenbauerC(2*n, -2*n, -1/2). - Peter Luschny, May 07 2016
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) = sqrt(-3)^(2*n)*P(2*n,-1/sqrt(-3)), where P(n,x) is the Legendre polynomia1 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)
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)
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. 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.
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EXAMPLE
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G.f. = 1 + 3*x + 19*x^2 + 141*x^3 + 1107*x^4 + 8953*x^5 + 73789*x^6 + ...
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MAPLE
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a := n -> simplify(GegenbauerC(2*n, -2*n, -1/2)):
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MATHEMATICA
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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 *)
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PROG
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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));
(Maxima) makelist(sum(binomial(2*n-k, k)*binomial(2*n, k), k, 0, n), n, 0, 40);
(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))
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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