No more terms below 10^8; 4294967297 is a term of this sequence.
5 first terms are Fermat primes (A019434).
Conjecture: next term is 4294967297.
Subsequence of A232720 and A000051.
Sequence is different from A232720 and A000215; A232720(6) = 83623937 and A000215(7) = 18446744073709551617 are not terms of this sequence.
From Jeppe Stig Nielsen, Nov 19 2016: (Start)
Since n1 is a solution to phi(x)=x/2, it is clear from the formula for phi that x=n1 is a nontrivial power of two (in A000079 and even). Then y=n2 is an odd number such that phi(y) is a power of two, and again recalling the formula for phi, this can only happen when y is a product of distinct Fermat primes (y in A045544).
Question: Is the above comment, together with Euler's demonstration that the Fermat number 2^32 + 1 = A000215(5) is composite, enough to prove that A262534 has only these six terms?
With the above observations, we need only search next to powers of two (A000051), so it is quick to determine that there are no terms between 65537 and 4294967297. (End)
From Robert Israel, Dec 08 2016: (Start)
The jth binary digit (i.e. coefficient of 2^j) of the product of a set of distinct Fermat numbers, say y=Product_{k in T} (2^(2^k)+1), is 1 iff j = Sum_{k in S} 2^k for some subset S of T. In order for x = y+1 to be a power of 2, all of y's binary digits must be 1. Since 2^(2^5)+1 is composite, T cannot contain 5, so digit 32 is 0, and y < 2^32. Thus we do have only these 6 terms. (End)
