cp's OEIS Frontend

A383966 Numbers k such that floor(2^k / 5) is a prime.

%I A383966 #18 Jun 08 2025 15:19:06
%S A383966 4,11,15,23,35,71,95,183,475,579,631,759,1519,1771,3031,6035,6951,
%T A383966 11423,37451,51935,68051
%N A383966 Numbers k such that floor(2^k / 5) is a prime.
%C A383966 From _David A. Corneth_, Jun 07 2025: (Start)
%C A383966 Observation from Hugo Pfoertner: For k > 4, terms are equivalent to 3 (mod 4).
%C A383966 Proof: We may write 2^k = 2^(4*m + r) = 16^m * 2^r = (15 + 1)^m * 2^r with 0 <= r <= 3.
%C A383966 Floor dividing by 5 gives 3*t + floor(2^r) for some positive integer t. floor(2^r) = 0 for r in {0, 1, 2} so then floor(2^k) is a multiple of 3. For k > 4 and r < 3 we definitely do get a multiple of 3 that is larger than 3 hence composite. (End)
%C A383966 a(22) > 150000. - _Hugo Pfoertner_, Jun 08 2025
%e A383966 From _David A. Corneth_, Jun 07 2025: (Start)
%e A383966 4 is in the sequence as floor(2^4/5) = 3 is prime.
%e A383966 5 is not in the sequence as floor(2^5/5) = 6 which is not prime.
%e A383966 7 is not in the sequence as floor(2^7/5) = 25 is not prime.
%e A383966 8 is not in the sequence as 8 > 4 and 8 is not equivalent to 3 (mod 4).
%e A383966 11 is in the sequence as floor(2^11/5) = 409 which is prime. (End)
%t A383966 Select[Range[1,20000],PrimeQ[Quotient[2^#,5]]&]
%o A383966 (Magma)  [n: n in [1..1000] | IsPrime(2^n div 5)];
%K A383966 nonn,more
%O A383966 1,1
%A A383966 _Vincenzo Librandi_, Jun 07 2025
%E A383966 a(19)-a(21) from _Hugo Pfoertner_, Jun 07 2025