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).

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

Author

Zhi-Wei Sun, Jul 25 2014

Keywords

• sign

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

Examples

			a(10) = -11 since L(2*10) = 15127 == -11 (mod prime(10)=29).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000032, A000040, A245525.

Programs

Mathematica

rMod[m_,n_]:=Mod[m,n,-(n-1)/2]
a[n_]:=rMod[LucasL[2n],Prime[n]]
Table[a[n],{n,1,80}]