OFFSET
1,1
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
FORMULA
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
EXAMPLE
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)
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Roger L. Bagula, Mar 02 2005
STATUS
approved