RE A007689: all terms composite?

[Tautócrona]

----- Original Message ----- 
From: "Tautócrona" <tautocrona at terra.es>

If we consider a(4k+2), then:

2^4 = = 3^4 = = 3 (mod 13)

2^(4k+2) + 3^(4k+2) = =
= = 3(4+9) = = 3·0 = = 0 (mod 13)

So a(4k+2) can never be prime.

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(I had a typo, it was 3^k·13, but the result is the same).

a) a(4*(4k+1)) = a(16k+4)

2^16 = = 3^16 = = 61 (mod 97)

a(16k+4) = = 61^k· 97 = = 0 (mod 97)

b) a(4*(4k+2)) = a(16k+8)

2^16 = = 3^16 = = 1 (mod 17)

a(16k+8) = = 6817 = = 0 (mod 17)

c) a(4*(4k+3)) = a(16k+12)

2^16 = = 3^16 = = 61 (mod 97)

a(16k+12) = = 535537 = = 0 (mod 97)

d) a(4*(4k)) = a(16k)

We see these residues don't have a simple formula, we need to separate again in special 
cases, namely a(4*(16k+1)), a(4*(16k+2)), a(4*16k+3)), a(4*16k). Of these, the first and 
the third are always divisible by 3041, and the second is always divisible by 1153, but 
the fourth hasn't a simple formula and then we need to separate again in four cases...

You see, it looks like a proccess we could prove by induction (a complicated one, though). 
I don't have the time now, so I will let it here. I'm pretty sure that there aren't primes 
for n>4.

Jose Brox