Hello, there are many exotic examples like that. Look at this one, the sequence A007699, a(n)= {nearest integer to } a(n-1)^2 /a(n-2). the sequence is : 10, 219, 4796, 105030, 2300104, 50371117, 1103102046, ... the ratio of a(n+1)/a(n) -->> 21.89949541893233438052532024370859294946 6190194047646611617481379446934015013... quite rapidly. By testing that number for algebraicity, one quickly finds that it is one of the real roots of 4 3 2 11 x - 18 x + 3 x - 22 x + 1 , the root can be calculated explicitely but it is quite ugly. BUT this is FALSE, the exact number, the ratio of a(n+1)/a(n) to high precision is : 21.89949541893233438052532024370859294946619019404764661\ 1617481379446934015013222134632935910397445152107249\ 9761349032053359109441276427149926005650693640825950\ 1335437931357406880506038288883182166121707555714731\ 7300725684058890389637589630893096102492114259366306\ 8931460123024765848662425871358354508021664249361199\ 6944188439509332536170556143293330518711985939602460\ 9252047620048792957280978376910436207360514815505096\ ... Is NOT a simple algebraic the root of the <apparent> equation is valid up to 0.11357748460267988639402531902 * 10^(-1878). As far as I know, nobody really knows WHY it behaves like that, those Pisot sequences are quite bizarre in my humble opinion. There are others, like A010900, that one too is exotic enough. E(4,13). A good idea would be to test against any asymptotics the sequence (21.8994954189323343805253202437085929494661)^n to see how far it is from the real sequence. I discussed that topic with David Boyd (UBC) a while ago and the problem with these animals is that we could have a situation where the exponents (or order) of the recurrence relation could be bery big. That means that we can have some approximations but the real thing is far more complicated. ps : I added this comment on the edit page of the sequence. PS2 : this is one example where whatever sophistication we now have with programs like PSLQ, LLL, Pari and all that, we still can find examples where it fails to give a definite answer. best regards, Simon Plouffe