A112683 For each positive integer n, consider the ternary sequence given initially by x(i) = 0 if 1 <= i < n, x(n) = 1; and thereafter determined by the quadratic recurrence x(i) = x(i-1) + x(i-n)^2 mod 3. Define a(n) to be the smallest positive integer N for which x(N+i) = x(i) for all sufficiently large i.

1, 4, 4, 9, 19, 4, 4, 22, 36, 4, 4, 45, 64, 4, 4, 102, 182, 213, 4, 188, 272, 4, 412, 225, 202, 4, 4, 1444, 512, 4, 4, 840, 1237, 4, 1138, 362, 1263, 4, 4, 1536, 672, 1786, 4, 701, 741, 4, 4, 2098, 3921, 5400, 178, 1183, 2348, 4, 7698, 6042, 5091, 4, 4

Offset

1,2

References

Terms computed by Bob Harder.

Links

Table of n, a(n) for n=1..59.

Steven R. Finch, Periodicity in Sequences Mod 3 [Broken link]

Steven R. Finch, Periodicity in Sequences Mod 3 [From the Wayback machine]

Example

For example, if n=4, then N=9, since the first 60 terms of x are:
0 0 0 1 1 1 1 2 0 1
2 0 0 1 2 2 2 0 1 2
0 0 1 2 2 2 0 1 2 0
0 1 2 2 2 0 1 2 0 0
1 2 2 2 0 1 2 0 0 1
2 2 2 0 1 2 0 0 1 2

Mathematica

period[lst_List] := Catch[lg = If[Length[lst] <= 5, 2, 40]; lst1 = lst[[1 ;; lg]]; km = Length[lst] - lg; Do[ If[lst1 == lst[[k ;; k+lg-1]], Throw[k-1]]; If[k == km, Throw[0]], {k, 2, km}]]; a[n_] := (ClearAll[x]; x[i_ /; 1 <= i < n] = 0; x[n] = 1; x[i_] := x[i] = Mod[x[i-1] + x[i-n]^2, 3]; xx = Table[x[i], {i, 1, 20000}]; period[xx // Reverse]); Table[a[n], {n, 1, 59}] (* Jean-François Alcover, Nov 30 2012 *)

Crossrefs

Cf. A112684, A112675.

Sequence in context: A214826 A135065 A067553 * A192030 A117879 A334009

Adjacent sequences: A112680 A112681 A112682 * A112684 A112685 A112686

Keyword

nonn

Author

N. J. A. Sloane, Dec 31 2005.

Status

approved