From April 29 to May 5, contributions came from Emmanuel Vantieghem, Adam
Stinchcombe and Oscar Volpatti
About puzzle 1129, Alan Rochelli was right, but
only for n>0.
Let's explicitly define his new
sequence b(n), with n>=0:
b(n) is the least prime such
that 3^(n+1), but not 3^(n+2), divides 2^(b(n)-1)-1.
It is possible to prove that b(n)
is also the least ODD prime of the form c*3^n+1 with c not
divisible by 3.
The proof uses a known theorem:
if g is a primitive root
modulo prime p, and if g^(p-1) is not congruent to 1 modulo p^2,
then g is also a primitive root modulo all powers p^k.
Number g=2 is a primitive root
modulo prime p=3;
congruence relation 2^2 == 4 !=
1 mod 9 holds too;
therefore 2 is also a primitive
root modulo all powers 3^k.
In other words, power 3^k divides
Mersenne number M(x) = 2^x-1 if and only if x itself is
divisible by Phi(3^k) = 2*3^(k-1).
Let's go back to sequence b(n).
3^(n+1) divides 2^(b(n)-1)-1,
so 2*3^n divides b(n)-1;
3^(n+2) doesn't divide 2^(b(n)-1)-1,
so 2*3^(n+1) doesn't divide b(n)-1;
therefore b(n)-1 = d*2*3^n, with d
not divisible by 3.
It must be d = 3*q+1 or d = 3*q+2,
so that b(n) has one of the following forms:
In both cases we have a linear
polynomial which produces primes for infinitely many values of
variable q, by Dirichlet's theorem on arithmetic progressions.
Let's compare Alan Rochelli's
sequence with OEIS A137990:
prime b(n) is of the form c*3^n+1 with c = 2*d, so that b(n)
is forced to be odd.
But, for n>0, a(n) is forced to be an odd prime too, because
it must be greater than prime 2:
a(n) >= 1*3^n+1 > 1*3^0+1 = 2.
Therefore identity a(n) = b(n) always
holds for n>0.
But it doesn't hold for
n = 0, because a(0) = 2 while b(0) = 3.