A245204 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.

1, 2, 2, 4, 1, 1, 5, 1, -2, -6, 10, 14, 5, 7, 7, -28, -12, 13, 14, 26, -21, -31, -13, -10, -11, -7, -6, 5, 2, -21, 2, 33, -15, -24, 34, 71, -15, 24, 9, 37, 73, -18, -84, -65, 9, -90, -65, -47, 97, -64, -100, -8, 41, 81, -81, -71, -65, -70, 113, 10, -80, 119, 57, 78, 20, 124, 167, -71, -48

Offset: 2

Author

Zhi-Wei Sun, Jul 13 2014

Keywords

• sign

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.

Examples

			a(3) = 2 since E_{prime(3)-3}(1/4) = E_2(1/4) = -3/16 == 2 (mod prime(3)=5).
		

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.

Crossrefs

Cf. A000040, A001586, A122045, A198245, A245089, A245206.

Programs

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}]