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
Keywords
Examples
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
References
- Terms computed by Bob Harder.
Links
- Steven R. Finch, Periodicity in Sequences Mod 3 [Broken link]
- Steven R. Finch, Periodicity in Sequences Mod 3 [From the Wayback machine]
Programs
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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 *)