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%I #42 Jun 20 2017 15:34:35
%S 29,41,89,281,1049,1048601,4194329,17179869209,1180591620717411303449,
%T 4951760157141521099596496921,5192296858534827628530496329220121,
%U 332306998946228968225951765070086169
%N Primes of the form 2^n + 5^2.
%C Primes of the form 4^n + 4! + 1. - _Vincenzo Librandi_, Nov 13 2010
%C Indeed, calculating mod 3 we have 2^n + 5^2 = (-1)^n + 1 = 0 if n is odd, so n must be even to yield a prime. - _M. F. Hasler_, Nov 13 2010
%C Those even values of n are given in A157006. Since n = 2k, these prime numbers also have the form 4^k + 25, where k is given in A204388. - _Timothy L. Tiffin_, Aug 06 2016
%C These primes a(m) can be used to generate numbers having deficiency 26. The formula a(m)*(a(m)-25)/2 produces those terms in A275702 having rightmost digit 8. - _Timothy L. Tiffin_, Aug 09 2016
%F a(m) = 2^(A157006(m)) + 5^2 = 4^(A204388(m)) + 25. - _Timothy L. Tiffin_, Aug 07 2016
%F If n == 0 mod 4, then a(m) == 1 mod 10. If n == 2 mod 4, then a(m) == 9 mod 10. - _Timothy L. Tiffin_, Aug 09 2016
%e From _Timothy L. Tiffin_, Aug 07 2016: (Start)
%e a(1) = 2^2 + 5^2 = 4 + 25 = 29.
%e a(2) = 2^4 + 5^2 = 16 + 25 = 41.
%e a(3) = 2^6 + 5^2 = 64 + 25 = 89.
%e a(4) = 2^8 + 5^2 = 256 + 25 = 281.
%e a(5) = 2^10 + 5^2 = 1024 + 25 = 1049.
%e a(6) = 2^20 + 5^2 = 1048576 + 25 = 1048601. (End)
%t a = Delete[Union[Flatten[Table[If [PrimeQ[2^n + 25] == True, 2^n + 25, 0], {n, 1, 400}]]], 1]
%t Select[2^Range[0,120]+25,PrimeQ] (* _Harvey P. Dale_, Jun 20 2017 *)
%Y Cf. A157006, A204388, A275702.
%K nonn
%O 1,1
%A _Roger L. Bagula_, Mar 02 2005