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A244741 Numbers k such that (prime(k) mod 5) == 2 (mod 3).

%I A244741 #20 Feb 04 2024 11:33:03
%S A244741 1,4,7,12,15,19,25,28,31,33,37,39,45,49,55,59,63,66,68,69,73,78,88,91,
%T A244741 93,101,102,106,107,111,113,118,123,129,134,138,139,144,148,151,154,
%U A244741 155,159,161,163,165,168,181,184,187,195,199,203,206,211,214,217
%N A244741 Numbers k such that (prime(k) mod 5) == 2 (mod 3).
%C A244741 Every positive integer is in exactly one of the sequences A244739, A024707, A244741.
%H A244741 Clark Kimberling, <a href="/A244741/b244741.txt">Table of n, a(n) for n = 1..1000</a>
%e A244741 n ... prime(n) mod 5 mod 3
%e A244741 1 ..... 2 ..... 2 ... 2
%e A244741 2 ..... 3 ..... 3 ... 0
%e A244741 3 ..... 5 ..... 0 ... 0
%e A244741 4 ..... 7 ..... 2 ... 2
%e A244741 5 ..... 11 .... 1 ... 1
%e A244741 6 ..... 13 .... 3 ... 0
%p A244741 A244741:=n->`if`(((ithprime(n) mod 5) mod 3) = 2, n, NULL): seq(A244741(n), n=1..250); # _Wesley Ivan Hurt_, Jul 06 2014
%t A244741 z = 300; u = Mod[Table[Mod[Prime[n], 5], {n, 1, z}], 3] (* A244738 *)
%t A244741 v1 = Flatten[Position[u, 0]]  (* A244739 *)
%t A244741 v2 = Flatten[Position[u, 1]]  (* A024707 *)
%t A244741 v3 = Flatten[Position[u, 2]]  (* A244741 *)
%Y A244741 Cf. A039703, A244738, A244739, A024707, A244735. Essentially the same as A049509.
%K A244741 nonn,easy
%O A244741 1,2
%A A244741 _Clark Kimberling_, Jul 05 2014