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%I #13 Jul 13 2014 17:09:14
%S 1,2,2,4,1,1,5,1,-2,-6,10,14,5,7,7,-28,-12,13,14,26,-21,-31,-13,-10,
%T -11,-7,-6,5,2,-21,2,33,-15,-24,34,71,-15,24,9,37,73,-18,-84,-65,9,
%U -90,-65,-47,97,-64,-100,-8,41,81,-81,-71,-65,-70,113,10,-80,119,57,78,20,124,167,-71,-48
%N The unique integer r with |r| < prime(n)/2 such that E_{prime(n)-3}(1/4) == r (mod prime(n)), where E_m(x) denotes the Euler polynomial of degree m.
%C Conjecture: a(n) = 0 infinitely often. In other words, there are infinitely many odd primes p such that E_{p-3}(1/4) == 0 (mod p) (equivalently, p divides A001586(p-3)).
%C This seems reasonable in view of the standard heuristic arguments. The first n with a(n) = 0 is 171 with prime(171) = 1019. The next such a number n is greater than 2600 and hence prime(n) > 23321.
%C Zhi-Wei Sun made many conjectures on congruences involving E_{p-3}(1/4), see the reference.
%H Zhi-Wei Sun, <a href="/A245204/b245204.txt">Table of n, a(n) for n = 2..1300</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1001.4453">Super congruences and Euler numbers</a>, arXiv:1001.4453 [math.NT].
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1007/s11425-011-4302-x">Super congruences and Euler numbers</a>, Sci. China Math. 54(2011), 2509-2535.
%e a(3) = 2 since E_{prime(3)-3}(1/4) = E_2(1/4) = -3/16 == 2 (mod prime(3)=5).
%t rMod[m_,n_]:=Mod[Numerator[m]*PowerMod[Denominator[m],-1,n],n,-n/2]
%t a[n_]:=rMod[EulerE[Prime[n]-3,1/4],Prime[n]]
%t Table[a[n],{n,2,70}]
%Y Cf. A000040, A001586, A122045, A198245, A245089, A245206.
%K sign
%O 2,2
%A _Zhi-Wei Sun_, Jul 13 2014
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