SUPERMULTIPLICATIVITY THEOREM: F(a+b) >= F(a) * F(b) because of the cartesian product. [Because of the lemma: If, among three among the 2-digit mixed radix numbers with bases F(a) and F(b), the first digits all are equal, then the second digits must also all be equal, or else not.] Hence, F(k) >= 2^k because F(1)=2 F(3*k) >= 9^k because F(3)=9 F(4*k) >= 20^k because F(4)=20 F(5*k) >= 45^k because F(5)=45 F(6*k) >= 112^k because F(6)=112 the lattermost yielding the growth constant 112^(1/6) = 2.1955... which is almost as good as the best known 2.21. The 2.21 presumably arises from F(11*k) >= 6464^k because F(11)>=6464 which apparently was shown by Yves Edel: Extensions of Generalized Product Caps, Designs, Codes and Cryptography 31,1 (January 2004) 5-14 http://link.springer.com/article/10.1023%2FA%3A102736590123120 but I have not seen this paper, so this is not certain. Actually 6464^(1/11) = 2.22028749761019184633 more precisely, so actually we could have said 2.22 not 2.21, if it really is true that F(11)>=6464, improving in the last decimal place over the lower bound F(n)>2.21^n that it says in the OEIS entry -- at least for all large-enough n. Another lemma: if you have a "SET" then adding any constant to all its digits (mod 3) still yields a SET. The constant can, in fact, depend on the digit location. Similarly you can multiply the Kth digit by any nonzero constant (mod 3). Also, you can permute the n digit locations using any fixed n-permutation. These generally yield many "equivalent versions" of the SET. Still more generally, you can transform all the n-vectors by multiplying by any invertible nXn constant matrix (mod 3), or adding on any constant n-vector (mod 3). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)