cp's OEIS Frontend

A104072 Primes of the form 2^n + 5^2.

%I A104072 #42 Jun 20 2017 15:34:35
%S A104072 29,41,89,281,1049,1048601,4194329,17179869209,1180591620717411303449,
%T A104072 4951760157141521099596496921,5192296858534827628530496329220121,
%U A104072 332306998946228968225951765070086169
%N A104072 Primes of the form 2^n + 5^2.
%C A104072 Primes of the form 4^n + 4! + 1. - _Vincenzo Librandi_, Nov 13 2010
%C A104072 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 A104072 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 A104072 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 A104072 a(m) = 2^(A157006(m)) + 5^2 = 4^(A204388(m)) + 25. - _Timothy L. Tiffin_, Aug 07 2016
%F A104072 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 A104072 From _Timothy L. Tiffin_, Aug 07 2016: (Start)
%e A104072 a(1) = 2^2  + 5^2 =       4 + 25 =      29.
%e A104072 a(2) = 2^4  + 5^2 =      16 + 25 =      41.
%e A104072 a(3) = 2^6  + 5^2 =      64 + 25 =      89.
%e A104072 a(4) = 2^8  + 5^2 =     256 + 25 =     281.
%e A104072 a(5) = 2^10 + 5^2 =    1024 + 25 =    1049.
%e A104072 a(6) = 2^20 + 5^2 = 1048576 + 25 = 1048601. (End)
%t A104072 a = Delete[Union[Flatten[Table[If [PrimeQ[2^n + 25] == True, 2^n + 25, 0], {n, 1, 400}]]], 1]
%t A104072 Select[2^Range[0,120]+25,PrimeQ] (* _Harvey P. Dale_, Jun 20 2017 *)
%Y A104072 Cf. A157006, A204388, A275702.
%K A104072 nonn
%O A104072 1,1
%A A104072 _Roger L. Bagula_, Mar 02 2005