It seems plausible that m exists for all n >= 0.
If a(n) = 2k, then a(n+1) <= k^2. If a(n) = 2k+1, then a(n+1) <= k*(k+1). Thus m exists for all n >= 0. -
Chai Wah Wu, Jun 15 2019
Conjecture: all terms, except for a(2), are either primes (
A000040) or squarefree semiprimes (
A006881). -
Chai Wah Wu, Jun 18 2019
If n != 1, then a(n+1) <= (a(n)-prevprime(a(n)))*prevprime(a(n)) where prevprime is
A151799.
Proof: Let m = (a(n)-prevprime(a(n)))*prevprime(a(n)). By Chebyshev's theorem (Bertrand's postulate), a(n)-prevprime(a(n)) <= prevprime(a(n)) and thus
A063655(m) = (a(n)-prevprime(a(n))) + prevprime(a(n)) = a(n). The only exception is when a(n) = 6. In this case m = 5, and
A308194(5) = 0 even though
A063655(5) = 6.
For n = 0, 2, 3, 11 and 17, this upper bound on a(n+1) is achieved, i.e., a(n+1) = (a(n)-prevprime(a(n)))*prevprime(a(n)).
Conjecture: a(n+1) = (a(n)-prevprime(a(n)))*prevprime(a(n)) infinitely often.
(End)
