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Primes of the form 2^n + 5^2.
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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