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

4, 11, 15, 23, 35, 71, 95, 183, 475, 579, 631, 759, 1519, 1771, 3031, 6035, 6951, 11423, 37451, 51935, 68051

Offset: 1

Author

Vincenzo Librandi, Jun 07 2025

Keywords

• nonn
• more

Comments

From David A. Corneth, Jun 07 2025: (Start)

Observation from Hugo Pfoertner: For k > 4, terms are equivalent to 3 (mod 4).

Proof: We may write 2^k = 2^(4*m + r) = 16^m * 2^r = (15 + 1)^m * 2^r with 0 <= r <= 3.

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)

a(22) > 150000. - Hugo Pfoertner, Jun 08 2025

Examples

			From _David A. Corneth_, Jun 07 2025: (Start)
4 is in the sequence as floor(2^4/5) = 3 is prime.
5 is not in the sequence as floor(2^5/5) = 6 which is not prime.
7 is not in the sequence as floor(2^7/5) = 25 is not prime.
8 is not in the sequence as 8 > 4 and 8 is not equivalent to 3 (mod 4).
11 is in the sequence as floor(2^11/5) = 409 which is prime. (End)
		

Programs

Magma

[n: n in [1..1000] | IsPrime(2^n div 5)];

Mathematica

Select[Range[1,20000],PrimeQ[Quotient[2^#,5]]&]

Extensions

a(19)-a(21) from Hugo Pfoertner, Jun 07 2025