[math-fun] derangements, three, shorter
conjecture (up to w=11) mod( !(3^w k + 1) , 3^(w+2) ) == 0 (w>1 ; all k>1) W. --- gibberish beyond this line --- =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.
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/
participants (2)
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Alec Mihailovs -
Meeussen Wouter (bkarnd)