A214834 a(n) is the smallest prime number such that both a(n) and (4n + 2) - a(n) are prime numbers of the form 4k + 3.

3, 3, 3, 7, 3, 3, 7, 3, 7, 11, 3, 3, 7, 11, 3, 7, 3, 3, 7, 3, 3, 7, 11, 19, 19, 3, 3, 7, 11, 19, 19, 3, 3, 7, 3, 7, 11, 3, 7, 11, 3, 3, 7, 11, 3, 7, 11, 3, 7, 3, 7, 11, 3, 7, 11, 3, 3, 7, 11, 3, 7, 11, 3, 7, 11, 3, 7, 3, 7, 11, 3, 7, 11, 47, 19, 23, 3, 3, 7, 11

Offset: 2

Author

Lei Zhou, Mar 07 2013

Keywords

• nonn
• easy

Comments

Any even number of the form of 4k + 2 (k > 0) can be partitioned into 4k + 2 = (4i+3) + (4j+3), where i + j + 1 = k. This sequence implies the conjecture that for any number in the form of 4k + 2 (k > 0), there is a partition for which 4i + 3 and 4j + 3 are both prime.

Conjecture tested true up to n=1000000000. In case the conjecture is not true, zero could be used to represent the missing entries.

Examples

			Let n = 4.  Then 4n + 2 = 14, and the pairs of prime numbers of the form 4k + 3 that sum to 14 are (3, 11), (7, 7).  The smallest number of 3, 11, 7, 8 is 3, so a(4) = 3.
		

Links

• Lei Zhou, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A005843, A008586, A016825, A000040, A002145.

Programs

Mathematica

s = 2; Table[s = s + 4; p1 = s + 1; While[p1 = p1 - 4; p2 = s - p1;  !((PrimeQ[p1]) && (PrimeQ[p2]) && (Mod[p2, 4] == 3))]; p2, {i, 1, 80}]