login
a(n) = least prime p such that 2*p + 1 equals (2*n - 1)*q where q is a prime, or 0 if no such p exists.
2

%I #25 Sep 08 2022 08:46:19

%S 0,7,7,17,13,71,19,37,59,47,31,149,37,67,43,263,181,227,1091,97,61,

%T 107,67,1433,73,127,79,137,199,383,701,157,97,167,103,461,109,487,269,

%U 197,283,2531,127,739,311,227,139,617,2861,643,151,257,157,1979,163,277

%N a(n) = least prime p such that 2*p + 1 equals (2*n - 1)*q where q is a prime, or 0 if no such p exists.

%C Conjecture: 7 <= a(n) <= 4*n^2 - 5*n + 1 for n > 1. This conjecture implies that for every odd k > 1 there exist two primes p and q < 2*k such that k = (2*p + 1)/q.

%C Every positive term belongs to A053176.

%H Arkadiusz Wesolowski, <a href="/A292084/b292084.txt">Table of n, a(n) for n = 1..10000</a>

%o (Magma) lst:=[]; for n in [2..56] do q:=1; repeat q+:=2; p:=Truncate((2*n*q-q-1)/2); until IsPrime(p) and IsPrime(q); Append(~lst, p); end for; [0] cat lst;

%o (PARI) a(n) = {if (n==1, return(0)); forprime(p=3, , q = (2*p+1)/(2*n-1); if ((denominator(q) == 1) && isprime(q), return (p)););} \\ _Michel Marcus_, Sep 16 2017

%Y Cf. A053176, A292083.

%K nonn

%O 1,2

%A _Arkadiusz Wesolowski_, Sep 08 2017