cp's OEIS Frontend

(2^2)*3-1=11, (2^3)*3-1=23 and (2^4)*3-1=47 are primes so 3 is the third entry.

For every x in A001122, the x-th term of this sequence and every succeeding term is divisible by x. For example 3 divides the 3rd and every succeeding term, 5 divides the 5th and every succeeding term.

The sequences of primes generated by these numbers are a type of Cunningham chain of the first kind (CC1). Since the longest known CC1 chain is of length 16, the next terms are currently unknown. - Douglas Stones (dssto1(AT)student.monash.edu.au), Mar 16 2005

These are pseudoprimes k == 3 (mod 4) such that 2k+1 is prime.

Proof. Let q = 2k+1 be prime, where k == 3 (mod 4) is a pseudoprime. We have q == 7 (mod 8), so 2 is a square mod q, which gives 2^((q-1)/2) == 1 (mod q), by Euler's criterion. Thus, 2^k == 1 (mod q), which implies 2^(k-1) == (q+1)/2 (mod q), so that 2^(k-1) == k+1 (mod q). The conclusion that 2^(k-1) == k+1 (mod kq) follows from the assumption that k is a pseudoprime and from the Chinese remainder theorem. - Carl Pomerance (in a letter to the author), Apr 14 2021

Note that if p is a Lucasian prime, i.e., p == 3 (mod 4) with 2p+1 prime; then (2^p-1)/(2p+1) == 1 (mod p), hence 2^p-2p-2 == 0 (mod p(2p+1)), so 2^(p-1) == p+1 (mod p(2p+1)).