A112683 - OEIS

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.

%I #13 Jul 09 2025 04:25:24

%S 1,4,4,9,19,4,4,22,36,4,4,45,64,4,4,102,182,213,4,188,272,4,412,225,

%T 202,4,4,1444,512,4,4,840,1237,4,1138,362,1263,4,4,1536,672,1786,4,

%U 701,741,4,4,2098,3921,5400,178,1183,2348,4,7698,6042,5091,4,4

%N 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.

%D Terms computed by Bob Harder.

%H Steven R. Finch, <a href="http://www.people.fas.harvard.edu/~sfinch/csolve/seqmod3.pdf">Periodicity in Sequences Mod 3</a> [Broken link]

%H Steven R. Finch, <a href="https://web.archive.org/web/20150911081920/http://www.people.fas.harvard.edu/~sfinch/csolve/seqmod3.pdf">Periodicity in Sequences Mod 3</a> [From the Wayback machine]

%e For example, if n=4, then N=9, since the first 60 terms of x are:

%e 0 0 0 1 1 1 1 2 0 1

%e 2 0 0 1 2 2 2 0 1 2

%e 0 0 1 2 2 2 0 1 2 0

%e 0 1 2 2 2 0 1 2 0 0

%e 1 2 2 2 0 1 2 0 0 1

%e 2 2 2 0 1 2 0 0 1 2

%t 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 *)

%Y Cf. A112684, A112675.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Dec 31 2005