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A245204
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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.
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2
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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
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2,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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a(3) = 2 since E_{prime(3)-3}(1/4) = E_2(1/4) = -3/16 == 2 (mod prime(3)=5).
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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