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A278405 a(n) = Sum_{k=0..n} binomial(n,2k)^2*binomial(n-k,k).

%I A278405 #20 Apr 24 2025 02:13:35
%S A278405 1,1,2,19,110,476,2477,15093,86830,485290,2826902,16857116,100034453,
%T A278405 594833357,3574477090,21611465819,130955824174,796195223398,
%U A278405 4860425688176,29760574848750,182655048136510,1123720751229858,6929124085148938,42811398244528788
%N A278405 a(n) = Sum_{k=0..n} binomial(n,2k)^2*binomial(n-k,k).
%C A278405 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 A278405 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.
%C A278405 Diagonal of the rational function 1 / ((1 + x)*(1 - x)*(1 - y)*(1 - z) - x*y*z). - _Ilya Gutkovskiy_, Apr 23 2025
%H A278405 Zhi-Wei Sun, <a href="/A278405/b278405.txt">Table of n, a(n) for n = 0..200</a>
%H A278405 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1610.03384">Supercongruences involving Lucas sequences</a>, arXiv:1610.03384 [math.NT], 2016.
%e A278405 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 A278405 G.f. = 1 + x + 2*x^2 + 19*x^3 + 110*x^4 + 476*x^5 + 2477*x^6 + 15093*x^7 + ...
%t A278405 a[n_]:=a[n]=Sum[Binomial[n,2k]^2*Binomial[n-k,k],{k,0,n/2}]
%t A278405 Table[a[n],{n,0,27}]
%Y A278405 Cf. A208425, A244973, A275027, A277640, A278415.
%K A278405 nonn
%O A278405 0,3
%A A278405 _Zhi-Wei Sun_, Nov 20 2016