A217696 Let p = A002145(n) be the n-th prime of the form 4k+3, then a(n) is the smallest number such that p is the smallest prime of the form 4k+3 for which 4*a(n)+2-p is prime.

1, 4, 10, 24, 76, 102, 196, 74, 104, 348, 314, 345, 86, 660, 443, 1494, 914, 1329, 2613, 1635, 1316, 1856, 1688, 2589, 2628, 6423, 3116, 2165, 6320, 4445, 7278, 4743, 16539, 17783, 6084, 3806, 6281, 8946, 15129, 6266, 10976, 19538, 16794, 31160, 32916, 57041

Offset: 1

Author

Lei Zhou, Mar 19 2013

Keywords

• nonn

Comments

It is conjectured that a(n) is defined for all positive integers.

This is also the index of first occurrence of the n-th prime in the form of 4k+3 in A214834.

Examples

			n=1: the first prime in the form of 4k+3 is 3, 3+3=6=4*1+2, so a(1)=1;
n=2: the second prime in the form of 4k+3 is 7, 7+7=14=3+11=4*3+2, and 11 is also a prime in the form of 4k+3, so a(2)!=3. 7+11=18=4*4+2=3+15, and 15 is not a prime number. So a(2)=4.
		

Links

• Lei Zhou, Table of n, a(n) for n = 1..170

Crossrefs

Cf. A214834, A016825, A000040, A002145, A155642.

Programs

Mathematica

goal = 46; plst = {}; pct = 0; clst = {}; n = -1; While[pct < goal,
n = n + 4; If[PrimeQ[n], AppendTo[plst, n]; AppendTo[clst, 0];
  pct++]]; n = 2; cct = 0; While[cct < goal, n = n + 4; p1 = n + 1;
While[p1 = p1 - 4; p2 = n - p1; ! ((PrimeQ[p1]) && (PrimeQ[p2]) && (Mod[p2, 4] == 3))]; If[MemberQ[plst, p2], If[id = Position[plst, p2][[1, 1]]; clst[[id]] == 0, clst[[id]] = (n - 2)/4; cct++]]]; clst

PARI

ok(n,p)=if(!isprime(n-p),return(0));forprime(q=2,p-1,if(q%4==3 && isprime(n-q),return(0)));1
a(n)=my(p,k); forprime(q=2,,if(q%4==3&&n--==0,p=q;break)); k=(p+1)/4; while(!ok(4*k+2,p),k++); k \\ Charles R Greathouse IV, Mar 19 2013