

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(i1) + x(in)^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.


2



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
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OFFSET

1,2


REFERENCES

Terms computed by Bob Harder.


LINKS

Table of n, a(n) for n=1..59.
S. R. Finch, Periodicity in Sequences Mod 3


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+lg1]], Throw[k1]]; 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[i1] + x[in]^2, 3]; xx = Table[x[i], {i, 1, 20000}]; period[xx // Reverse]); Table[a[n], {n, 1, 59}] (* JeanFrançois Alcover, Nov 30 2012 *)


CROSSREFS

Cf. A112684, A112675.
Sequence in context: A214826 A135065 A067553 * A192030 A117879 A069549
Adjacent sequences: A112680 A112681 A112682 * A112684 A112685 A112686


KEYWORD

nonn


AUTHOR

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


STATUS

approved



