[seqfan] Re: (3^n+1)/2 is prime, A171381
Robert Israel
israel at math.ubc.ca
Sun Jun 13 08:51:05 CEST 2010
Note that if j is odd, (3^(jn) + 1)/2 is divisible by (3^n+1)/2, so
all terms of the sequence must be powers of 2. Thus
A171831(n) = 2^A138083(n).
Robert Israel israel at math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
On Sun, 13 Jun 2010, N. J. A. Sloane wrote:
> Jon S. corrected this sequence. After 9 incorrect terms
> were deleted, it reads:
>
> %I A171381
> %S A171381 1,2,4,16,32,64
> %N A171381 Numbers n such that (3^n + 1)/2 is prime.
> %C A171381 Note that n must be a power of 2.
> %C A171381 The next term, if it exists, is at least 2^15. - Jon Schoenfield, Jun 12 2010.
> %e A171381 For n = 1 we obtain a(1) = (3^1 + 1)/2 = 2 which is a prime.
> %Y A171381 Cf. A138083.
> %Y A171381 Sequence in context: A173746 A095803 A032464 this_sequence A145119 A081411 A094384
> %Y A171381 Adjacent sequences: A171378 A171379 A171380 this_sequence A171382 A171383 A171384
> %K A171381 nonn,more,new
> %O A171381 1,2
> %A A171381 Joao Carlos Leandro da Silva (zxawyh66(AT)yahoo.com), Dec 07 2009
> %E A171381 Edited by N. J. A. Sloane, Dec 09 2009
> %E A171381 Incorrect terms a(7)-a(15) deleted Jun 12 2010 by Jon Schoenfield
>
> (or in other words, 0,1,2,4,5,6,>=15)
>
> Can anyone find the next term? (Otherwise it is awfully close to the
> powers of 2)
>
> Neil
>
>
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