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#27 by Harvey P. Dale at Wed Mar 23 14:40:20 EDT 2022
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#26 by Harvey P. Dale at Wed Mar 23 14:40:17 EDT 2022
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| MATHEMATICA
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Position[Accumulate[2^(2*Range[1000]-1)], _?(PrimeQ[#+1]&)]//Flatten (* The program generates the first 21 terms of the sequence. To generate more, increase the Range constant. *). *) (* _Harvey P. Dale_, Mar 23 2022 *)
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| STATUS
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approved
editing
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#25 by Harvey P. Dale at Wed Mar 23 14:39:30 EDT 2022
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#24 by Harvey P. Dale at Wed Mar 23 14:39:26 EDT 2022
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| MATHEMATICA
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Position[Accumulate[2^(2*Range[1000]-1)], _?(PrimeQ[#+1]&)]//Flatten (* The program generates the first 21 terms of the sequence. To generate more, increase the Range constant. *)
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approved
editing
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#23 by N. J. A. Sloane at Mon Sep 24 16:53:14 EDT 2018
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| EXTENSIONS
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Edited by N. J. A. Sloane at the suggestion of _Andrew S. Plewe_, Jun 11 2007
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Discussion
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Mon Sep 24
| 16:53
| OEIS Server: https://oeis.org/edit/global/2767
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#22 by Bruno Berselli at Wed Sep 16 06:13:38 EDT 2015
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#21 by Michel Marcus at Wed Sep 16 05:59:11 EDT 2015
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#20 by Jon E. Schoenfield at Wed Sep 16 05:56:37 EDT 2015
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#19 by Jon E. Schoenfield at Wed Sep 16 05:56:35 EDT 2015
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| NAME
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Numbers n such that 1 + Sum_{i=1..n} [} 2^(2i-1)] ) is prime.
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| COMMENTS
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Terms beyond a(30) correspond to probable primes, cf. A000978. [From _. - _M. F. Hasler_, Aug 29 2008]
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a(n) = floor[ (A000978(n)/2 ] = ceil( ) = ceiling(log[(4]()(A000979(n))) ; ))); A000978(n) = 2 a(n) + 1 ; ; A000979(n) = (2*4^a(n)+1)/3. [From _. - _M. F. Hasler_, Aug 29 2008]
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| PROG
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(PARI) for(n=1, 999, ispseudoprime(2^(2*n+1)\3+1) & print1(n", ")) \ [From _", ")) \\ _M. F. Hasler_, Aug 29 2008]
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| EXTENSIONS
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Edited by N. J. A. Sloane at the suggestion of _Andrew Plewe, _, Jun 11 2007
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| STATUS
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approved
editing
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#18 by N. J. A. Sloane at Sat Sep 06 00:47:22 EDT 2014
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