%I #22 May 11 2020 01:38:54
%S 1,25,20,30,24,20
%N Number of 5 X 5 permutation matrices such that the n-th matrix power is the least nonnegative power that gives the identity matrix.
%C The sum of the terms of this sequence is equal to the number of 5 X 5 permutation matrices: 5! = 120.
%C Number of elements of order n in symmetric group S_5. - _Alois P. Heinz_, Mar 30 2020
%e a(1) = 1 because there is only one matrix whose first power is the identity matrix (this is the identity matrix itself).
%t tab = {0, 0, 0, 0, 0, 0}; per =
%t Permutations[{1, 2, 3, 4, 5}]; zeromat = {}; Do[
%t AppendTo[zeromat, Table[0, {n, 1, 5}]], {m, 1, 5}]; unit =
%t IdentityMatrix[5]; s5 = {}; Do[s = zeromat;
%t Do[s[[m]][[per[[n]][[m]]]] = 1, {m, 1, 5}];
%t AppendTo[s5, s], {n, 1, 120}]; Do[
%t If[MatrixPower[s5[[n]], 1] == unit, tab[[1]] = tab[[1]] + 1,
%t If[MatrixPower[s5[[n]], 2] == unit, tab[[2]] = tab[[2]] + 1,
%t If[MatrixPower[s5[[n]], 3] == unit, tab[[3]] = tab[[3]] + 1,
%t If[MatrixPower[s5[[n]], 4] == unit, tab[[4]] = tab[[4]] + 1,
%t If[MatrixPower[s5[[n]], 5] == unit, tab[[5]] = tab[[5]] + 1,
%t If[MatrixPower[s5[[n]], 6] == unit,
%t tab[[6]] = tab[[6]] + 1]]]]]], {n, 1, 120}]; tab
%Y Row n=5 of A057731.
%K nonn,fini,full
%O 1,2
%A _Artur Jasinski_, Dec 18 2009
%E Name edited and terms corrected by _Alois P. Heinz_, Mar 30 2020