A223489 a(n) = number of missing residues in the Lucas sequence mod the n-th prime number.

0, 0, 1, 0, 4, 1, 1, 7, 4, 19, 12, 9, 22, 10, 32, 9, 22, 33, 16, 27, 17, 30, 20, 65, 17, 66, 24, 74, 61, 73, 30, 49, 37, 106, 77, 114, 33, 40, 40, 49, 67, 119, 72, 49, 49, 183, 181, 54, 56, 149, 205, 90, 138, 94, 61, 178, 149, 102, 73, 254, 70, 81, 264, 117, 69

Offset: 1

Author

Casey Mongoven, Mar 20 2013

Keywords

• nonn

Comments

The Lucas numbers mod n for any n are periodic - see A106291 for period lengths.

Examples

			The 5th prime number is 11. The Lucas sequence mod 11 is {2,1,3,4,7,0,7,7,3,10,2,1,3,...} - a periodic sequence. There are 4 residues which do not occur in this sequence, namely {5,6,8,9}. So a(5) = 4.
		

References

• V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A137751.

Programs

Mathematica

pisano[n_] := Module[{a = {2, 1}, a0, k = 0, s}, If[n == 1, 1, a0 = a; Reap[While[k++; s = Mod[Plus @@ a, n]; Sow[s]; a[[1]] = a[[2]]; a[[2]] = s; a != a0]][[2, 1]]]]; Join[{2}, Table[u = Union[pisano[n]]; mx = Max[u]; Length[Complement[Range[0,mx], u]], {n, Prime[Range[2, 100]]}]] (* T. D. Noe, Mar 22 2013 *)