|
|
A233521
|
|
Number of disjoint subsets s of 0..(n-1) such that, for every x in s, x^x (mod n) is in s.
|
|
3
|
|
|
1, 1, 1, 2, 1, 4, 2, 4, 3, 5, 1, 7, 2, 5, 7, 7, 3, 9, 2, 10, 8, 7, 3, 13, 5, 10, 5, 13, 3, 15, 4, 11, 9, 10, 9, 15, 2, 7, 12, 19, 6, 20, 4, 12, 15, 7, 4, 22, 11, 16, 12, 15, 2, 16, 14, 18, 10, 9, 1, 30, 7, 8, 22, 19, 16, 21, 4, 17, 9, 23, 4, 27, 5, 10, 19, 14, 14
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
COMMENTS
|
This is very loosely based on the work of Kurlberg et al. It appears that a(n) = 1 at only six n: 1, 2, 3, 5, 11, 59.
|
|
LINKS
|
|
|
EXAMPLE
|
The simplest nontrivial case is n = 4. In this case, a(4) = 2 because there are two subsets: {0,1,2} and {3}. Note that 0^0 == 1 (mod 4), 1^1 == 1 (mod 4), 2^2 == 0 (mod 4), and 3^3 == 3 (mod 4).
|
|
MATHEMATICA
|
Table[toDo = Range[0, n-1]; sets = {}; While[Length[toDo] > 0, k = toDo[[1]]; toDo = Rest[toDo]; lst = {k}; While[q = PowerMod[k, k, n]; ! MemberQ[lst, q], AppendTo[lst, q]; toDo = Complement[toDo, {q}]; k = q]; AppendTo[sets, lst]]; Do[int = Intersection[sets[[i]], sets[[j]]]; If[int != {}, sets[[i]] = Union[sets[[i]], sets[[j]]]; sets[[j]] = {}], {i, Length[sets]}, {j, i+1, Length[sets]}]; Length[DeleteCases[sets, {}]], {n, 100}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|