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%I #13 Oct 06 2015 03:51:02
%S 0,0,2,-1,-2,-1,7,-5,-5,11,10,-11,11,12,2,17,-2,19,-15,-26,33,17,-22,
%T -11,18,8,18,-27,17,51,-37,-34,28,-4,66,-37,-69,-58,45,-81,-20,-86,
%U -19,17,-12,-30,35,-32,5,-11,-8,-45,12,-111,-28,-71,76,59,102,-25
%N Unique integer r with -prime(n)/2 < r <= prime(n)/2 such that S(n) == r (mod prime(n)), where S(n) is the large Schroeder number A006318(n).
%C Conjecture: (i) For any integer n > 2, the term a(n) is nonzero, i.e., prime(n) does not divide the large Schroeder number S(n).
%C (ii) For any integer n > 2, prime(n) does not divide the Bell number B(2*n) = A000110(2*n).
%C We have verified parts (i) and (ii) for n up to 440000 and 66000 respectively.
%C Conjecture (i) fails for the first time for n=20239789. In particular, a(20239789)=0. - _Max Alekseyev_, Oct 05 2015
%H Zhi-Wei Sun, <a href="/A245595/b245595.txt">Table of n, a(n) for n = 1..5000</a>
%e a(5) = -2 since S(5) = 394 == -2 (mod prime(5)=11).
%t rMod[m_,n_]:=Mod[m,n,-(n-1)/2]
%t S[n_]:=Sum[Binomial[n+k,2k]*Binomial[2k,k]/(k+1),{k,0,n}]
%t a[n_]:=rMod[S[n],Prime[n]]
%t Table[a[n],{n,1,60}]
%Y Cf. A000040, A000110, A006318, A245525.
%K sign
%O 1,3
%A _Zhi-Wei Sun_, Jul 27 2014
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