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A189286 a(n):=(Sum_{k=0}^n C(6k,3k)C(3k,k)C(6(n-k),3(n-k))C(3(n-k),n-k))/((2n-1)Binomial[3n,n]) 1


%S -1,40,696,23408,969496,44602560,2187147600,111957721920,

%T 5911097451480,319469892415808,17584481176101952,982222958294603040,

%U 55530668360895219728,3171318959654377396864,182670436050532943578560,10599737781026193970325760,619014530633087163062727000,36353266320338484003053582400,2145559172529803104937217263040,127190916635938933740168015020160

%N a(n):=(Sum_{k=0}^n C(6k,3k)C(3k,k)C(6(n-k),3(n-k))C(3(n-k),n-k))/((2n-1)Binomial[3n,n])

%C On Apr 19 2011, _Zhi-Wei Sun_ conjectured that a(n) is an integer for every n=0,1,2,.... He proved that a(p-1)=[(p+1)/6] (mod p) for any prime p, and also made the following conjecture:

%C (i) a(n)^{1/n} tends to 64 as n tends to the infinity.

%C (ii) For any positive integer n, we have a(n)=0 (mod 8), and a(n)/8 is odd if and only if n is a power of two.

%F Recursion: (n+2)^2*(3n+2)(3n+4)(3n+5)a(n+2)

%F =16(2n+1)(2n+3)(3n+2)(18n^2+54n+41)a(n+1) - 9216(n+1)^2(4n^2-1)(3n+5)a(n).

%e For n=1 we have a(1)=(C(6,3)C(3,1)+C(6,3)C(3,1))/C(3,1)=120/3=40.

%t S[n_]:=Sum[Binomial[6k,3k]Binomial[3k,k]Binomial[3(n-k),n-k]Binomial[6(n-k),3(n-k)],{k,0,n}]/((2n-1)Binomial[3n,n])

%t Table[S[n],{n,0,19}]

%K sign

%O 0,2

%A _Zhi-Wei Sun_, Apr 19 2011

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