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

29, 41, 89, 281, 1049, 1048601, 4194329, 17179869209, 1180591620717411303449, 4951760157141521099596496921, 5192296858534827628530496329220121, 332306998946228968225951765070086169

Offset: 1

Author

Roger L. Bagula, Mar 02 2005

Keywords

• nonn

Comments

Primes of the form 4^n + 4! + 1. - Vincenzo Librandi, Nov 13 2010

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

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

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

Examples

			From _Timothy L. Tiffin_, Aug 07 2016: (Start)
a(1) = 2^2  + 5^2 =       4 + 25 =      29.
a(2) = 2^4  + 5^2 =      16 + 25 =      41.
a(3) = 2^6  + 5^2 =      64 + 25 =      89.
a(4) = 2^8  + 5^2 =     256 + 25 =     281.
a(5) = 2^10 + 5^2 =    1024 + 25 =    1049.
a(6) = 2^20 + 5^2 = 1048576 + 25 = 1048601. (End)
		

Crossrefs

Cf. A157006, A204388, A275702.

Programs

Mathematica

a = Delete[Union[Flatten[Table[If [PrimeQ[2^n + 25] == True, 2^n + 25, 0], {n, 1, 400}]]], 1]
Select[2^Range[0,120]+25,PrimeQ] (* Harvey P. Dale, Jun 20 2017 *)

Formula

a(m) = 2^(A157006(m)) + 5^2 = 4^(A204388(m)) + 25. - Timothy L. Tiffin, Aug 07 2016

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