

A284285


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


0




OFFSET

1,1


COMMENTS

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


LINKS

Table of n, a(n) for n=1..10.
V. Shevelev, Compact integers and factorials, Acta Arithmetica 126 (2007), no. 3, 195236.


FORMULA

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


MATHEMATICA

Select[Range[7500], PrimeQ[(2^# + 3)/5] &] (* Michael De Vlieger, Mar 24 2017 *)


PROG

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


CROSSREFS

Cf. A000040, A283657.
Sequence in context: A147074 A147192 A259665 * A025624 A147496 A147383
Adjacent sequences: A284282 A284283 A284284 * A284286 A284287 A284288


KEYWORD

nonn,more


AUTHOR

Vladimir Shevelev, Mar 24 2017


EXTENSIONS

Terms a(4)a(9) from Peter J. C. Moses, Mar 24 2017
a(10) from Giovanni Resta, Mar 24 2017


STATUS

approved



