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A173774
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The arithmetic mean of (21*k + 8)*binomial(2*k,k)^3 (k=0..n-1).
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13
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8, 120, 3680, 144760, 6427008, 306745824, 15364514880, 796663553400, 42395640372800, 2302336805317120, 127078484504270208, 7108177964254013920, 402042028998035350400, 22954860061516225396800
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OFFSET
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1,1
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COMMENTS
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On Feb 10 2010, Zhi-Wei Sun introduced the sequence and conjectured that each term a(n) is an integer divisible by 4*binomial(2*n,n). On Feb 11 2011, _Kasper Andersen_ confirmed this conjecture by noting that the sequence b(n) = a(n)/(4*binomial(2*n,n)), for n > 0, coincides with A112029. It was proved that for every prime p and positive integer a we have a(p^a) == 8 + 1*6*p^3*B_(p-3) (mod p^4), where B_0, B_1, B_2, ... are Bernoulli numbers. Given a prime p, it has been conjectured that Sum_{k=0..(p-1)/2} (21*k + 8)*binomial(2*k,k)^3 == 8*p + (-1)^((p-1)/2)*32*p^3*E_(p-3) (mod p^4) if p > 3 (where E_0, E_1, E_2, ... are Euler numbers), and that Sum_{k=0..floor(2p^a/3)} (21*k + 8)*binomial(2*k,k)^3 == 8*p^a (mod p^(a + 5 + (-1)^p)) if a is a positive integer with p^a == 1 (mod 3). He also observed that b(n) = a(n)/(4*binomial(2*n,n)) is odd if and only if n is a power of two.
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LINKS
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FORMULA
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a(n) = (1/n)*Sum_{k=0..n-1} (21*k + 8)*binomial(2*k,k)^3.
(n+1)*a(n+1) = n*a(n) + 8*(21*n + 8)*binomial(2*n-1, n)^3, n > 0, with a(1) = 8.
a(n) = (1/n)*Sum_{j=0..n-1} (21*j + 8)*(j+1)^3*Catalan(j)^3. - G. C. Greubel, Jul 06 2021
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EXAMPLE
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For n=2 we have a(2)=120 since (8*binomial(0,0)^3 + (21+8)*binomial(2,1)^3)/2 = 120.
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MATHEMATICA
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a[n_]:= Sum[(21*k+8)*Binomial[2*k, k]^3, {k, 0, n-1}]/n; Table[a[n], {n, 25}]
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PROG
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(Magma) [(&+[(21*j+8)*(j+1)^3*Catalan(j)^2: j in [0..n-1]])/n: n in [1..30]]; // G. C. Greubel, Jul 06 2021
(Sage) [(1/n)*sum((21*j+8)*binomial(2*j, j)^3 for j in (0..n-1)) for n in (1..30)] # G. C. Greubel, Jul 06 2021
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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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