definition of anti-divisor

[Max Alekseyev]

Furthermore, since 2^(k+1)*p-1, 2^(k+1)*p+1 must equal -1 and +1
modulo 3, the number 2^(k+1)*p must be 0 modulo 3, implying that p=3.

Therefore, every element of A066466, except 4, must be of the form
3*2^k such that 3*2^(k+1)-1, 3*2^(k+1)+1 are twin primes. There no
such new k (i.e., except known 1,2,6,18) below 1000.

Max

On 7/23/07, Max Alekseyev <maxale at gmail.com> wrote:
> Except element 4, the elements of A066466 have form 2^k*p where p is
> odd prime and both 2^(k+1)*p-1, 2^(k+1)*p+1 are prime (i.e., twin
> primes).
> In other words, A066466 without element 4 is a subsequence of A040040,
> containing elements of the form 2^k*p with prime p.
>
> Max
>
> On 7/23/07, Jonathan Post <jvospost3 at gmail.com> wrote:
> > Any work since 2001 on whether or not there is a 6th anti-prime?  How
> > far has this been searched?  Any proofs or disproofs as to finiteness
> > of A066466?
> >
> > COMMENT FROM Jonathan Vos Post RE A066466
> >
> > %I A066466
> > %S A066466 3, 4, 6, 96, 393216
> > %N A066466 Numbers having just one anti-divisor.
> > %C A066466 Jon Perry calls these anti-prime numbers, saying that these
> > are the only 5 known. This sequence is worth extending, if possible,
> > or proving finite.
> > %F A066466 A066272(a(n)) = 1.
> > %Y A066466 Cf. A066272.
> > %O A066466 1
> > %K A066466 ,more,nonn,
> > %A A066466 Jonathan Vos Post (jvospost2 at yahoo.com), Jul 23 2007
> >
>