

A283191


Prime numbers p > 2 such that (2^p  5)/3 is prime.


0




OFFSET

1,1


COMMENTS

Let W = (2^p  5)/3 and s = (W+1)/(2*p), then 5^s == 2 (mod W) for terms 1..9.
Subsequence of 7, 13, 19, 31, 51, 55, 85, 111, 319, 373,.. which are numbers m such that (2^m5)/3 is prime.  R. J. Mathar, Mar 05 2017


LINKS

Table of n, a(n) for n=1..9.


MATHEMATICA

Select[Prime@ Range[2, 1000], PrimeQ[(2^#  5)/3] &] (* Michael De Vlieger, Mar 03 2017 *)


PROG

(PARI)
forprime(p=3, 30000, W= (2^p5)/3; if(ispseudoprime(W), print1(p, ", ")))


CROSSREFS

Cf. A000978.
Sequence in context: A023255 A122482 A265629 * A048375 A198035 A208720
Adjacent sequences: A283188 A283189 A283190 * A283192 A283193 A283194


KEYWORD

nonn,more


AUTHOR

Dmitry Ezhov, Mar 02 2017


STATUS

approved



