From: "Meeussen Wouter (bkarnd)" <wouter.meeussen@vandemoortele.com>
conjecture (up to w=11)
mod( !(3^w k + 1) , 3^(w+2) ) == 0 (w>1 ; all k>1)
Here is the proof. First, ----------------------------------------- Lemma. !(3k+1) = 0 mod 9 for all k>=0. ----------------------------------------- Proof. By induction. f(1)=0 - true. Let it be true for k=n-1, i.e. !(3n-2) = 0 mod 9. Then !(3n+1)=3n*(!(3n)+!(3n-1))=3n*((3n-1)*(!(3n-1)+!(3n-2))+!(3n-1))=9n^2*!(3n-1)+3n*(3n-1)*!(3n-2) =0 mod 9, q.e.d. ----------------------------------------- ----------------------------------------- Theorem. !(3^w k+1) = 0 mod 3^(w+2) for w>1, all k>=0. ----------------------------------------- Proof. Using the recurrence from the Lemma's proof, !(3^w k+1)=3^(2w) k^2*!(3^w k - 1) + 3^w k*(3^w k - 1)*!(3^w k - 2). The first term is divisible by 3^(w+2) because 2w=w+w>=w+2 for w>=2. The second term is divisible by 3^(w+2) because !(3^w k - 2) is divisible by 9 according to Lemma. Q.E.D. ------------------------------------------ Alec Mihailovs http://math.tntech.edu/alec/