2
$\begingroup$

$a_n,b_n$ are two sequence valued in $[0,1]$ and $a_0=1,b_0=0 $.

the following equation holds: $$a_n=\sum_{k=1}^{n}b_ka_{n-k}\tag{1}$$

$$A=\{n:a_n>0\}-\{0\}$$$$B=\{n:b_n>0\}$$ further more,assume $A$ is a addition semigroup and$$\text{g.c.d}(A)=1\tag{2}$$ then how to derive: $$\text{g.c.d}(B)=1\tag{3}$$


My teacher said that $A$ is generated by $B$,so they have the same g.c.d.,I don't know why.

$\endgroup$
  • $\begingroup$ If $\gcd(B)=d\ne1$, so $d$ divides $b_k$ for all $k$, then $d$ divides $a_n$ for all $n$, no? $\endgroup$ – Gerry Myerson Dec 12 '14 at 10:12
  • $\begingroup$ @GerryMyerson thanks,you are right.I asked such a stupid question:D $\endgroup$ – Lookout Dec 12 '14 at 13:29
  • $\begingroup$ Not to worry, we all overlook things now and then. $\endgroup$ – Gerry Myerson Dec 12 '14 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.