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Prime numbers n such that (2^n+1)/3 can be written in the form a^2+7*b^2.
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%I #7 Sep 04 2020 14:30:05

%S 5,7,11,13,17,19,23,31,43,53,61,67,79,89,101,127,139,167,179,191,199,

%T 211,223,227,269,313,317,347,389,431,433,457,503,593,601,613,619,673,

%U 701,739,797,827,839,907,967

%N Prime numbers n such that (2^n+1)/3 can be written in the form a^2+7*b^2.

%C These numbers of the form (2^n+1)/3 have no prime factors congruent to 3 (mod 7) or 5 (mod 7) or 6 (mod 7) to an odd power. Also the power of 2 is not 1.

%H Samuel S. Wagstaff, Jr. <a href="http://homes.cerias.purdue.edu/~ssw/cun/index.html">The Cunningham Project</a>

%t Select[Prime[Range[170]],FindInstance[a^2+7b^2==(2^#+1)/3,{a,b},Integers] != {}&] (* _Harvey P. Dale_, Sep 04 2020 *)

%K nonn

%O 1,1

%A _V. Raman_, Sep 08 2012