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A244469
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a(0) = 0, thereafter, a(n) = 2^(2*n-1)*( binomial((3*n-1)/2,n) - binomial(3*n/2, n)/3 ).
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2
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0, 1, 7, 58, 515, 4746, 44758, 428772, 4154403, 40599130, 399429602, 3950996556, 39255152846, 391466112324, 3916110379020, 39281346256008, 394942611929379, 3978982062756090, 40160256911157610, 405995113593507900, 4110284071450416090, 41666530928566504620, 422876855107176561780
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OFFSET
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0,3
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LINKS
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FORMULA
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G.f.: g'(x)/g(x)-1, g(x)=(2*sqrt(9*x+1)*sin(arcsin((54*x^2+27*x+2)/(2*(9*x+1)^(3/2)))/3))/3-1/3. - Vladimir Kruchinin, Apr 14 2019
a(n) = (1/n)*Sum_{k = 0..n} k*2^(n-k)*binomial(n+k-1,k)*binomial(2*n-k-1,n-k) for n >= 1.
a(n) = [x^n] G(x)^n = [x^n] 1/(1 - x*C(2*x))^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108 and G(x) is the g.f. of A064062.
n*(n-1)*(6*n-7)*a(n) = - 18*(n-1)*a(n-1) + 12*(3*n-5)*(6*n-1)*(3*n-4)*a(n-2) with a(1) = 1 a(2) = 7.
exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + x + 4*x^2 + 23*x^3 + 156*x^4 + ... is the g.f. of A007297.
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. (End)
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MAPLE
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f4:=n->-(2^(2*n-1)/3)*binomial(3*n/2, n) + 2^(2*n-1)*binomial((3*n-1)/2, n);
[seq(f4(n), n=1..40)]; # then prepend f4(0)=0.
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MATHEMATICA
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Join[{0}, Table[-(2^(2 n - 1)/3) Binomial[3 n/2, n] + 2^(2 n - 1) Binomial[(3 n - 1)/2, n], {n, 1, 30}]] (* Vincenzo Librandi, Jun 29 2014 *)
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PROG
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(PARI) {a(n) = if(n==0, 0, 2^(2*n-1)*(binomial((3*n-1)/2, n) - binomial(3*n/2, n)/3) )}; \\ G. C. Greubel, Apr 17 2019
(Magma) [n eq 0 select 0 else Round(2^(2*n-1)*(Gamma((3*n+1)/2)/Gamma((n+1)/2) - Gamma((3*n+2)/2)/(3*Gamma((n+2)/2)))/Factorial(n)): n in [0..30]]; // G. C. Greubel, Apr 17 2019
(Sage)
def a(n):
if n==0: return 0
else: return 2^(2*n-1)*(binomial((3*n-1)/2, n) - binomial(3*n/2, n)/3)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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