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%I #16 Apr 08 2018 03:24:35
%S 2,3,7,59
%N Numbers such that (2^(n^2)-1)/(2^n-1) is prime.
%C It is easy to see that all terms of this sequence must be prime; this motivates the definition of A051156(n) = (2^prime(n)^2-1)/(2^prime(n)-1).
%C No further terms up to n=1999. - _Andreas Höglund_, Apr 06 2018
%t Select[Prime[Range[17]], PrimeQ[Cyclotomic[#^2, 2]] &] (* _Arkadiusz Wesolowski_, May 13 2012 *)
%o (PARI) for/*prime*/( n=1,99, is/*pseudo*/prime( (2^n^2-1)/(2^n-1) ) & print1(n,","))
%Y Cf. A051156.
%K hard,more,nonn
%O 1,1
%A _M. F. Hasler_, Feb 10 2009
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