Re my first one--"period 8"/3^(n/4)^2. I assume he won't mind being quoted to the List, since he simply took me at my word, if a bit heavy-handedly: WM>----------------------------------------------------- I don't get it: you say it's 3^(n^2/16) time a (fixed) period 8 but I find 3^n {1,1,-1,0,1,-1,-1,0} to be the denominator of: {s[1],s[2],s[3],s[4]}={1,1,-1,-2/3}; s[n_]:=s[n]=(s[n-1]*s[n-3]+s[n-2]^2)/s[n-4] /. Indeterminate->0 s/@ Range[48] FactorInteger[%]//ColumnForm or am I reading you all haywireishly? Wouter. ------------------------------------------------------<MW I.e., he got 1, 1, -1, -2/3, 1/3, 1/9, -1/27, 0, 1/243, -1/729, -1/2187, 0, 1/19683, 1/59049, -1/177147, 0, 1/1594323, -1/4782969, -1/14348907, 0,... whereas I dodged 0/0 by plugging x=-2/3 into 1, 1, -1, x, 1 + x, -1 - x + x^2, -1 - x - x^3,... getting 1, 1, -1, -2/3, 1/3, 1/9, -1/27, 0, 1/243, -1/729, -1/2187, 2/19683, 1/59049, -1/531441, -1/4782969, 0, I guess Wouter should have been a little more cautious about 0/0. --rwg