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a(n) = Sum_{k=0..n} binomial(n,2k)^2*binomial(n-k,k).
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%I #16 Sep 26 2019 05:51:32

%S 1,1,2,19,110,476,2477,15093,86830,485290,2826902,16857116,100034453,

%T 594833357,3574477090,21611465819,130955824174,796195223398,

%U 4860425688176,29760574848750,182655048136510,1123720751229858,6929124085148938,42811398244528788

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

%C Conjecture: For any prime p > 5 and positive integer n, the number (a(p*n)-a(n))/(p*n)^3 is always a p-adic integer.

%C We have proved that for any prime p > 5 and positive integer n the number (a(p*n)-a(n))/(p^3*n^2) is always a p-adic integer.

%H Zhi-Wei Sun, <a href="/A278405/b278405.txt">Table of n, a(n) for n = 0..200</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1610.03384">Supercongruences involving Lucas sequences</a>, arXiv:1610.03384 [math.NT], 2016.

%e a(3) = 19 since a(3) = C(3,2*0)^2*C(3-0,0) + C(3,2*1)^2*C(3-1,1) = 1 + 3^2*2 = 19.

%e G.f. = 1 + x + 2*x^2 + 19*x^3 + 110*x^4 + 476*x^5 + 2477*x^6 + 15093*x^7 + ...

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

%t Table[a[n],{n,0,27}]

%Y Cf. A208425, A244973, A275027, A277640, A278415.

%K nonn

%O 0,3

%A _Zhi-Wei Sun_, Nov 20 2016