

A223489


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


1



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
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OFFSET

1,5


COMMENTS

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


REFERENCES

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


LINKS



EXAMPLE

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.


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 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



