OFFSET
0,1
COMMENTS
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
From Chai Wah Wu, Jun 22 2019: (Start)
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)
LINKS
Rémy Sigrist, C program for A308195
PROG
(C) See Links section.
CROSSREFS
KEYWORD
nonn,more
AUTHOR
N. J. A. Sloane, Jun 14 2019
EXTENSIONS
a(14)-a(20) from Rémy Sigrist, Jun 14 2019
a(21) from Chai Wah Wu, Jun 17 2019
a(22) from Chai Wah Wu, Jun 24 2019
a(23) from Giovanni Resta, Jun 25 2019
STATUS
approved