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A245526
Unique integer r with -prime(n)/2 < r <= prime(n)/2 such that L(2*n) == r (mod prime(n)), where L(k) denotes the Lucas number A000032(k).
2
1, 1, -2, -2, 2, -3, -7, 3, 5, -11, -15, 8, -18, -14, 3, -12, 19, -18, 25, 14, 5, 21, 11, 7, -22, 3, 43, -40, -7, -53, 54, 23, 11, -12, -7, 41, 6, -13, -66, 71, -32, 18, 94, -20, -79, 7, -88, 12, 11, -73, 3, 29, -120, 50, 10, -60, -63, 34, 94, 47, -113, 131, -18, 128, 60, 57, 79, 22, -45, -68, 100, 100, 131, -171, 56, -166, 11, -153, -174, 10
OFFSET
1,3
COMMENTS
Conjecture: a(n) is always nonzero, i.e., prime(n) never divides the Lucas number L(2*n).
We have verified this for all n = 1, ..., 2*10^6.
On Jul 26 2014, Bjorn Poonen (from MIT) found a counterexample with n = 14268177. - Zhi-Wei Sun, Jul 26 2014
EXAMPLE
a(10) = -11 since L(2*10) = 15127 == -11 (mod prime(10)=29).
MATHEMATICA
rMod[m_, n_]:=Mod[m, n, -(n-1)/2]
a[n_]:=rMod[LucasL[2n], Prime[n]]
Table[a[n], {n, 1, 80}]
CROSSREFS
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Jul 25 2014
STATUS
approved