A284285 Numbers k such that (2^k + 3)/5 is prime.

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

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.

a(11) > 160000. - Michael S. Branicky, Jul 12 2024

Links

Table of n, a(n) for n=1..10.

V. Shevelev, Compact integers and factorials, Acta Arithmetica 126 (2007), no. 3, 195-236.

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,changed

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