A308195 a(n) = smallest m such that A308194(m) = n, or -1 if no such m exists.

5, 6, 8, 7, 10, 11, 23, 29, 101, 137, 757, 1621, 3238, 15537, 44851, 155269, 784522, 2495326, 7485969, 51719803, 119775247, 2017072213, 5629349191, 40094417851

Offset: 0

Author

N. J. A. Sloane, Jun 14 2019

Keywords

• nonn
• more

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

Crossrefs

Cf. A063655, A308194, A151799.

Programs

C

See Links section.

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