|
%I #17 Mar 13 2013 13:24:04
%S 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,
%T 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,
%U 7,11,3,7,11,3,7,3,7,11,3,7,11,47,19,23,3,3,7,11
%N 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.
%C 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.
%C Conjecture tested true up to n=1000000000. In case the conjecture is not true, zero could be used to represent the missing entries.
%H Lei Zhou, <a href="/A214834/b214834.txt">Table of n, a(n) for n = 2..10000</a>
%e 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.
%t 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}]
%Y Cf. A005843, A008586, A016825, A000040, A002145.
%K nonn,easy
%O 2,1
%A _Lei Zhou_, Mar 07 2013
|
