A119980 Order of the following permutation on 3n+1 symbols. Write the 3n+1 symbols horizontally into a 3-column grid and read them off vertically, i.e., column after column.

1, 3, 6, 6, 11, 15, 52, 38, 51, 9, 360, 260, 35, 39, 364, 1932, 680, 532, 1122, 260, 2415, 3570, 168, 360, 71, 12285, 836, 12, 1680, 1155, 858, 936, 7956, 48300, 171120, 234, 4428, 235752, 712, 990, 119, 364182, 406, 11220, 412920, 25584, 476, 19998, 6486

Offset: 0

Author

Roland Miyamoto, Aug 03 2006

Keywords

• nonn

Examples

			For n=2, the grid with 0..6 by rows is
   0 1 2
   3 4 5      first column is one longer
   6
Reading them by columns gives (0,3,6,1,4,2,5) which as a permutation has order 6, so a(2) = 6.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1000

Crossrefs

The case for 2 columns is A002326.

Cf. A003572.

Programs

GAP

# GAP / KANT / KASH
# SpartaEncrypt(d,L) returns the list M obtained by writing L in d columns
# and then concatenating these columns
SpartaEncrypt := function(d,L)
local len, i, M;
len := Length(L);
M := [];
for i in [1..d] do
Append(M,L{[i,d+i..d*IntQuo(len-i,d)+i]});
od;
return M;
end;
# SpartaOrd(d,m) computes the order of SpartEncrypt(d,[0..m-1])
# in the group S_m
SpartaOrd := function(d,m)
local L, M, i;
M := [0..m-1];
L := [0..m-1];
i := 0;
repeat
i := i + 1;
L := SpartaEncrypt(d,L);
until L=M;
return i;
end;
d:=3; r:=1;
a := List([0..60],n->SpartaOrd(d,d*n+r));

PARI

P(n,w,j)={my(c=j%w); if(c==0, j/w, j\w + c*n + 1)}
Follow(s,f)={my(t=f(s), k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)}
CyclePoly(n,w,x)={my(q=0); for(i=0, w*n, my(l=Follow(i, j->P(n,w,j))); if(l, q+=x^l)); q}
a(n)={my(q=CyclePoly(n, 3, x), m=1); for(i=1, poldegree(q), if(polcoef(q, i), m=lcm(m, i))); m} \\ Andrew Howroyd, Jan 04 2024