%I #2 Mar 30 2012 17:37:58

%S 1,1,4,1,16,4,9,1,36,16,100,4,144,9,16,1,64,36,324,16,36,100,121,4,

%T 400,144,324,9

%N Parse the base-2 expansion of 1/n using the Ziv-Lempel encoding as described in A106182; sequence gives the eventual period of the differences of the sequence of lengths of the successive phrases.

%C The Ziv-Lempel encoding scans the sequence from left to right and inserts a comma when the current phrase (since the last comma) is distinct from all previous phrases (between commas).

%C It appears that a(n) is just the square of the period of the base 2 expansion of 1/n. For example, if n=3 the sequence of terms in the base-2 expansion of 1/3 is {0,1,0,1,0,1,0,1,...}, of period 2, whereas a(3)=4=2^2.

%e For n=3, the sequence of base-2 digits of 1/3 is {0,1,0,1,0,1,0,1,0,1,0,1,...}. The Ziv-Lempel encoding parses this into "phrases": {0}, {1}, {0,1}, {0,1,0}, {1,0}, {1,0,1}, {0,1,0,1}, {0,1,0,1,0}, {1,0,1,0}, {1,0,1,0,1}, {0,1,0,1,0,1}, ..., with lengths {1,1,2,3,2,3,4,5,4,5,6,7,6,7,8,9,8,9,10,11,...}. The differences are {0,1,1,-1,1,1,1,-1,1,1,1,-1,1,...} which quickly becomes periodic with period 4. Thus a(3)=4.

%Y Cf. A106182, A109337.

%K nonn

%O 1,3

%A _John W. Layman_, Sep 24 2010