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

0,1

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

Table of n, a(n) for n=0..23.

Rémy Sigrist, C program for A308195

Prog

(C) See Links section.

Crossrefs

Cf. A063655, A308194, A151799.

Sequence in context: A011498 A330863 A111514 * A308191 A111770 A105738

Adjacent sequences: A308192 A308193 A308194 * A308196 A308197 A308198

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