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a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n-k,k)^2 * 81^k.
1

%I #10 Nov 09 2019 01:11:51

%S 1,1,325,2917,247861,5937301,265793401,10705726585,378746444917,

%T 18588932910901,657940881863305,32580334626782185,1257522211980656425,

%U 59212895251349313865,2490039488311462939645,112553667120196462181437

%N a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n-k,k)^2 * 81^k.

%C On Jul 17 2010, _Zhi-Wei Sun_ introduced this sequence and made the following conjecture: If p is a prime with p=1,9,11,19 (mod 40) and p = x^2+10y^2 with x,y integers, then Sum_{k=0..p-1} a(k) == 4x^2-2p (mod p^2); if p is a prime with p == 7,13,23,37 (mod 40) and 2p = x^2 + 10y^2 with x,y integers, then Sum_{k=0..p-1} a(k) == 2p - 2x^2 (mod p^2); if p is an odd prime with (-10/p)=-1, then Sum_{k=0..p-1} a(k) == 0 (mod p^2). He also conjectured that Sum_{k=0..n-1} (10k+9)*a(k) == 0 (mod n) for all n=1,2,3,... and that Sum_{k=0..p-1} (10k+9)*a(k) == p(4(-2/p)+5) (mod p^2) for any prime p > 3.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/0911.5665">Open Conjectures on Congruences</a>, preprint, arXiv:0911.5665 [math.NT], 2009-2011.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1006.2776">On Apery numbers and generalized central trinomial coefficients</a>, preprint, arXiv:1006.2776 [math.NT], 2010-2011.

%e For n=2 we have a(2) = 1 + 2^2*81 = 325.

%t a[n_]:=Sum[Binomial[n,k]^2Binomial[n-k,k]^2*81^k,{k,0,n}] Table[a[n],{n,0,25}]

%Y Cf. A179524, A178790, A178791, A178808, A173774.

%K nonn

%O 0,3

%A _Zhi-Wei Sun_, Jul 18 2010