OFFSET
2,2
COMMENTS
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)).
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.
Zhi-Wei Sun made many conjectures on congruences involving E_{p-3}(1/4), see the reference.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 2..1300
Zhi-Wei Sun, Super congruences and Euler numbers, arXiv:1001.4453 [math.NT].
Zhi-Wei Sun, Super congruences and Euler numbers, Sci. China Math. 54(2011), 2509-2535.
EXAMPLE
a(3) = 2 since E_{prime(3)-3}(1/4) = E_2(1/4) = -3/16 == 2 (mod prime(3)=5).
MATHEMATICA
rMod[m_, n_]:=Mod[Numerator[m]*PowerMod[Denominator[m], -1, n], n, -n/2]
a[n_]:=rMod[EulerE[Prime[n]-3, 1/4], Prime[n]]
Table[a[n], {n, 2, 70}]
CROSSREFS
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Jul 13 2014
STATUS
approved