A007689: all terms composite?

[zak seidov]

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 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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