OFFSET
3,1
COMMENTS
Conjecture: a(n) = 0 infinitely often. In other words, there are infinitely many primes p > 3 such that B_{p-2}(1/3) == 0 (mod p).
This seems reasonable in view of the standard heuristic arguments. Our computation shows that if a(n) = 0 then n > 2600 and hence prime(n) > 23000.
Zhi-Wei Sun made many conjectures on congruences involving B_{p-2}(1/3), see the Sci. China Math. paper and arXiv:1407.0967.
The first value of n with a(n) = 0 is 18392. For the prime p = prime(18392) = 205129, we have B_{p-2}(1/3) == 20060*p (mod p^2). - Zhi-Wei Sun, Dec 13 2014
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 3..1300
Guo-Shuai Mao and Zhi-Wei Sun, Two congruences involving harmonic numbers with applications, arXiv:1412.0523 [math.NT], 2014.
Zhi-Wei Sun, Super congruences and Euler numbers, Sci. China Math. 54(2011), 2509-2535.
Zhi-Wei Sun, Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k, arXiv:1407.0967 [math.NT], 2014.
EXAMPLE
a(3) = -2 since B_{prime(3)-2}(1/3) = B_3(1/3) = 1/27 == -2 (mod prime(3)=5).
MATHEMATICA
rMod[m_, n_]:=Mod[Numerator[m]*PowerMod[Denominator[m], -1, n], n, -n/2]
a[n_]:=rMod[BernoulliB[Prime[n]-2, 1/3], Prime[n]]
Table[a[n], {n, 3, 70}]
CROSSREFS
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Jul 11 2014
STATUS
approved