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%I #26 Nov 17 2021 08:15:57
%S 1,1,5,19,85,401,1931,9605,48469,248365,1286605,6726875,35441275,
%T 187935775,1002122525,5369287019,28889315669,156015203845,
%U 845330354321,4593724615175,25029614166685,136704935601785,748273234994675,4103928115592365,22549175326327675,124105065258631651,684100888645922051,3776354280849020005
%N a(n) = Sum_{k=0..n} C(n,k)^2*C(n-k,k), where C(n,k) denotes the binomial coefficient n!/(k!*(n-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 The author has 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 As a(n) = Sum_{k=0..n} C(n,k)*C(n,2k)*C(2k,k) and C(2k,k) = 2*C(2k-1,k-1) for k = 1,2,3,..., we see that a(n) is always odd. We guess that a(n) is congruent to one of 0, 1, -1 modulo 5.
%H Zhi-Wei Sun, <a href="/A275027/b275027.txt">Table of n, a(n) for n = 0..200</a>
%H Joel A. Henningsen and Armin Straub, <a href="https://arxiv.org/abs/2111.08641">Generalized Lucas congruences and linear p-schemes</a>, arXiv:2111.08641 [math.NT], 2021.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1610.03384">Supercongruences involving Lucas sequences</a>, arXiv:1610.03384 [math.NT], 2016.
%F a(n) = Sum_{k=0..n}C(n,k)*C(n,2k)*C(2k,k).
%F By the Zeilberger algorithm, we have the recurrence (n+3)^2*(23n+25)*a(n+3) = 25*(n+1)^2*(23n+48)*a(n) + (391n^3+1989n^2+3288n+1750)*a(n+1) + (46n^3+280n^2+ 519n+265)*a(n+2) for all n >= 0.
%F a(n) = hypergeom([-n, 1/2 - n/2, -n/2], [1, 1], -4). - _Peter Luschny_, Mar 21 2018
%F a(n) ~ c * d^n / (Pi*n), where d = 5.729031537980930837932235459792820714... is the real root of the equation -25 - 17*d - 2*d^2 + d^3 = 0 and c = 1.107089291883984657933126801836156175486638498732... is the positive real root of the equation -125 + 1048*c^2 - 2576*c^4 + 1472*c^6 = 0. - _Vaclav Kotesovec_, Jun 09 2019
%e a(2) = 5 since a(2) = Sum_{k=0,1,2}C(2,k)^2*C(2-k,k) = C(2,0)^2*C(2,0) + C(2,1)^2*C(1,1) = 1 + 4 = 5.
%t a[n_]:=a[n]=Sum[Binomial[n,k]^2*Binomial[n-k,k],{k,0,n/2}]
%t Table[a[n],{n,0,27}]
%t a[n_] := HypergeometricPFQ[{-n, 1/2 - n/2, -n/2}, {1, 1}, -4];
%t Table[a[n], {n, 0, 27}] (* _Peter Luschny_, Mar 21 2018 *)
%o (PARI) a(n) = sum(k=0, n, binomial(n,k)^2*binomial(n-k,k)); \\ _Michel Marcus_, Nov 13 2016
%Y Cf. A000984, A005258, A208425, A244973, A277640.
%K nonn
%O 0,3
%A _Zhi-Wei Sun_, Nov 12 2016
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