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 A308195 a(n) = smallest m such that A308194(m) = n, or -1 if no such m exists. 2
 5, 6, 8, 7, 10, 11, 23, 29, 101, 137, 757, 1621, 3238, 15537, 44851, 155269, 784522, 2495326, 7485969, 51719803, 119775247, 2017072213, 5629349191, 40094417851 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified September 26 02:27 EDT 2021. Contains 347664 sequences. (Running on oeis4.)