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Numbers k such that gcd(k, prime(k) + prime(k+1)) = 1.
16

%I #18 Apr 21 2021 03:49:30

%S 1,5,7,9,11,13,17,19,21,23,25,29,31,37,39,41,43,47,49,51,53,59,61,67,

%T 71,73,77,79,83,85,89,91,95,97,99,101,103,107,109,111,113,115,119,121,

%U 123,125,127,129,131,133,137,139,145,149,151,155,157,161,163

%N Numbers k such that gcd(k, prime(k) + prime(k+1)) = 1.

%C This sequence and A336367 partition the positive integers.

%e In the following table, p(k) = A000040(k) = prime(k).

%e k p(k) p(k)+p(k+1) gcd

%e 1 2 5 1

%e 2 3 8 4

%e 3 5 12 3

%e 4 7 18 2

%e 5 11 24 1

%e 6 13 30 6

%e Thus 1 and 5 are in this sequence; 2 and 3 are in A336367; 2 and 11 are in A336368; 3 and 5 are in A336369.

%t p[n_] := Prime[n];

%t u = Select[Range[200], GCD[#, p[#] + p[# + 1]] == 1 &] (* A336366 *)

%t v = Select[Range[200], GCD[#, p[#] + p[# + 1]] > 1 &] (* A336367 *)

%t Prime[u] (* A336368 *)

%t Prime[v] (* A336369 *)

%o (PARI) isok(m) = gcd(m, prime(m)+prime(m+1)) == 1; \\ _Michel Marcus_, Jul 20 2020

%Y Cf. A000040, A336367, A336368, A336369.

%K nonn

%O 1,2

%A _Clark Kimberling_, Jul 19 2020