%I #23 Jul 12 2024 21:59:08

%S 5,9,25,45,61,189,289,489,7065,42229

%N Numbers k such that (2^k + 3)/5 is prime.

%C For primes p = (2^a(n) + 3)/5, n >= 2, there are exactly 5 positive integers m for which the exponents of 2 and p in the prime power factorization of m! are both powers of 2 (for a proof see the Shevelev link (Theorem 2)).

%C The sequence of such primes p begins 7, 103, 6710887, 7036874417767, etc.

%C a(11) > 160000. - _Michael S. Branicky_, Jul 12 2024

%H V. Shevelev, <a href="http://journals.impan.gov.pl/aa/Inf/126-3-1.html">Compact integers and factorials</a>, Acta Arithmetica 126 (2007), no. 3, 195-236.

%F a(n) == 1 (mod 4).

%t Select[Range[7500], PrimeQ[(2^# + 3)/5] &] (* _Michael De Vlieger_, Mar 24 2017 *)

%o (PARI) is(n)=n%4==1 && isprime((2^n+3)/5) \\ _Charles R Greathouse IV_, Mar 25 2017

%Y Cf. A000040, A283657.

%K nonn,more

%O 1,1

%A _Vladimir Shevelev_, Mar 24 2017

%E Terms a(4)-a(9) from _Peter J. C. Moses_, Mar 24 2017

%E a(10) from _Giovanni Resta_, Mar 24 2017