A007689: all terms composite?

[zak seidov]

Note that this part of Michael Somos' comment
in %C A082101 
isn't OK(?):
>2^m+3^m is divisible by 2+3=5 unless m is zero
> or a power of 2. 

zak  



--- zak seidov <zakseidov at yahoo.com> wrote:

> I found this in OEIS:
> %I A082101
> %S A082101 2,5,13,97
> %N A082101 Primes of form 2^n+3^n.
> %C A082101 Next term, if it exists, is > 10^125074.
> -
> David Wasserman (wasserma(AT)spawar.navy.mil), Aug
> 13
> 2004
> %C A082101 Since x+y is a factor of x^m+y^m if m is
> odd, 2^m+3^m is divisible by 2+3=5 unless m is zero
> or a
> power of 2. This is similar to Fermat numbers 1+2^m.
> -
> Michael Somos, Aug 27 2004
> %e A082101 m=0: 1+1, m=1: 2+3, m=2: 4+9, m=4: 16+81
> %Y A082101 Cf. A094474-A094499.
> %Y A082101 Sequence in context: A075742 A075737
> A100843 this_sequence A090472 A065797 A016737
> %Y A082101 Adjacent sequences: A082098 A082099
> A082100
> this_sequence A082102 A082103 A082104
> %K A082101 more,nonn
> %O A082101 1,1
> %A A082101 Labos E. (labos(AT)ana1.sote.hu), Apr 14
> 2003
> 
> 
> --- Dean Hickerson <dean at math.ucdavis.edu> wrote:
> 
> > zak seidov asked when a(n) = 2^n + 3^n is prime:
> > 
> > > a(n) are primes for n=0,1,2,4.
> > > My Q: what about other terms a(2k), 
> > > are they all composite?
> > 
> > This is similar to the case of Fermat primes.  If
> k
> > is odd, then a(nk) is
> > divisible by a(n), since:
> > 
> >     a(nk) = (2^n)^k + (3^n)^k
> > 
> >           = (2^n + 3^n) [(2^n)^(k-1) - (2^n)^(k-2)
> > (3^n) + - ... + (3^n)^(k-1)]
> > 
> > So the only possible primes in the sequence are
> a(0)
> > and a(2^n) for n>=1.
> > I've checked that a(2^n) is composite for 3 <= n
> <=
> > 15.  As with Fermat
> > primes, a probabilistic argument suggests that
> there
> > are only finitely many
> > primes in the sequence, but I doubt that anyone
> can
> > prove it.
> > 
> > Dean Hickerson
> > dean at math.ucdavis.edu
> > 
> 
> 
> 
> 		
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