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%I #12 Nov 03 2019 16:28:04
%S 1,1,0,-5,-16,-24,15,197,576,724,-1200,-8832,-22801,-21293,76440,
%T 408795,922368,499104,-4446588,-19025060,-37012416,-1673992,245604832,
%U 880263936,1441226991,-908700649,-13088509200,-40222012703,-52991533744,88167061704,678172355415,1805175708261,1747974632448,-6237554623536,-34300087628480
%N a(n) = Sum_{k=0..n} binomial(n, 2k)*binomial(n-k, k)*(-1)^k.
%C Conjecture: For any prime p > 3 and positive integer n, the number (a(p*n)-a(n))/(p*n)^2 is always a p-adic integer.
%C We are able to show that for any prime p > 3 and positive integer n the number (a(p*n)-a(n))/(p^2*n) is always a p-adic integer.
%C See also A275027 and A278405 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A278415/b278415.txt">Table of n, a(n) for n = 0..400</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) = -5 since a(3) = C(3, 2*0)*C(3-0, 0)(-1)^0 + C(3,2*1)*C(3-1,1)(-1)^1 = 1 - 6 = -5.
%t a[n_]:=Sum[Binomial[n,2k]Binomial[n-k,k](-1)^k,{k,0,n}]
%t Table[a[n],{n,0,34}]
%Y Cf. A208425, A244973, A277640, A275027, A278405.
%K sign
%O 0,4
%A _Zhi-Wei Sun_, Nov 21 2016
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