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.

-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

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

Cf. A027641, A027642.

Sequence in context: A026190 A262599 A160016 * A335919 A296340 A048213

Adjacent sequences: A245086 A245087 A245088 * A245090 A245091 A245092

Keyword

sign

Author

Zhi-Wei Sun, Jul 11 2014

Status

approved