

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.


2



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
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OFFSET

2,2


COMMENTS

Conjecture: a(n) = 0 infinitely often. In other words, there are infinitely many odd primes p such that E_{p3}(1/4) == 0 (mod p) (equivalently, p divides A001586(p3)).
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.
ZhiWei Sun made many conjectures on congruences involving E_{p3}(1/4), see the reference.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 2..1300
ZhiWei Sun, Super congruences and Euler numbers, arXiv:1001.4453 [math.NT].
ZhiWei Sun, Super congruences and Euler numbers, Sci. China Math. 54(2011), 25092535.


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

Cf. A000040, A001586, A122045, A198245, A245089, A245206.
Sequence in context: A243487 A143485 A181633 * A099320 A206714 A230442
Adjacent sequences: A245201 A245202 A245203 * A245205 A245206 A245207


KEYWORD

sign


AUTHOR

ZhiWei Sun, Jul 13 2014


STATUS

approved



