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A117739
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Decimal expansion of the largest C_0 = 1.2209864... such that for C < C_0 and A < 2 the sequence a(n) = floor[A^(C^n)] can't contain only prime terms.
(history;
published version)
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#18 by Charles R Greathouse IV at Mon Apr 03 10:36:11 EDT 2023
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Chris K. Caldwell, <a href="httphttps://primes.utmt5k.eduorg/notes/proofs/A3n.html">A proof of a generalization of Mills' Theorem</a>
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Discussion
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Mon Apr 03
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| OEIS Server: https://oeis.org/edit/global/2966
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#17 by OEIS Server at Thu Jun 12 21:04:30 EDT 2014
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| LINKS
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Andrey V. Kulsha, <a href="/A117739/b117739_1.txt">Table of n, a(n) for n = 1..50000</a>
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#16 by N. J. A. Sloane at Thu Jun 12 21:04:30 EDT 2014
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Discussion
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Thu Jun 12
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| OEIS Server: Installed new b-file as b117739.txt. Old b-file is now b117739_1.txt.
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#15 by N. J. A. Sloane at Thu Jun 12 21:04:27 EDT 2014
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| COMMENTS
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It is not provenproved that for C > C_0 the mentioned infinite sequence of primes actually exists. However, heuristics show that A243358 could be infinite (the decimal expansion of corresponding A value is A243370).
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proposed
editing
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#14 by Andrey V. Kulsha at Thu Jun 12 14:38:25 EDT 2014
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#13 by Andrey V. Kulsha at Thu Jun 12 14:37:36 EDT 2014
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| LINKS
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Chris K. Caldwell, <a href="http://primes.utm.edu/notes/proofs/A3n.html">A proof of a generalization of Mills' Theorem</a>
Chris K. Caldwell, <a href="http://primes.utm.edu/notes/proofs/A3n.html">A proof of a generalization of Mills' Theorem</a>
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#12 by Andrey V. Kulsha at Thu Jun 12 14:36:38 EDT 2014
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| LINKS
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Andrey V. Kulsha, <a href="/A117739/b117739.txt">Table of n, a(n) for n = 1..10300</a>
Andrey V. Kulsha, <a href="/A117739/b117739_1.txt">Table of n, a(n) for n = 1..50000</a>
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proposed
editing
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#11 by Andrey V. Kulsha at Thu Jun 12 13:18:27 EDT 2014
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#10 by Andrey V. Kulsha at Thu Jun 12 13:18:05 EDT 2014
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| CROSSREFS
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Cf. A243358, sequence (primes), A243370 (value of primesA), A051021 (Mills' constant)
Cf. A243370, value of A
Cf. A051021, Mills' constant, where floor[Mills^(3^n)] is always prime
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proposed
editing
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Discussion
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Thu Jun 12
| 13:18
| Andrey V. Kulsha: Done.
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#9 by Michel Marcus at Thu Jun 12 13:09:28 EDT 2014
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