

A245089


The unique integer r with r < prime(n)/2 such that B_{prime(n)2}(1/3) == r (mod prime(n)), where B_m(x) denotes the Bernoulli polynomial of degree m.


6



2, 1, 4, 6, 8, 6, 10, 5, 3, 16, 4, 6, 3, 6, 11, 29, 2, 7, 21, 4, 16, 23, 5, 43, 14, 3, 32, 26, 13, 23, 64, 52, 30, 74, 17, 33, 37, 82, 68, 55, 78, 96, 79, 22, 81, 26, 7, 70, 38, 9, 3, 118, 128, 123, 67, 69, 78, 138, 30, 60, 19, 88, 26, 110, 27, 63, 82, 138
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OFFSET

3,1


COMMENTS

Conjecture: a(n) = 0 infinitely often. In other words, there are infinitely many primes p > 3 such that B_{p2}(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.
ZhiWei Sun made many conjectures on congruences involving B_{p2}(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_{p2}(1/3) == 20060*p (mod p^2).  ZhiWei Sun, Dec 13 2014


LINKS

ZhiWei Sun, Table of n, a(n) for n = 3..1300
GuoShuai Mao and ZhiWei Sun, Two congruences involving harmonic numbers with applications, arXiv:1412.0523 [math.NT], 2014.
ZhiWei Sun, Super congruences and Euler numbers, Sci. China Math. 54(2011), 25092535.
ZhiWei 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

Cf. A027641, A027642.
Sequence in context: A026190 A262599 A160016 * A296340 A048213 A283309
Adjacent sequences: A245086 A245087 A245088 * A245090 A245091 A245092


KEYWORD

sign


AUTHOR

ZhiWei Sun, Jul 11 2014


STATUS

approved



