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Primes p(n) such that gcd(n, prime(n)+prime(n+2)) = 1.
5

%I #4 Oct 18 2020 22:36:25

%S 2,5,11,17,31,41,47,59,67,83,103,109,127,149,157,167,179,191,211,227,

%T 241,257,277,283,307,313,331,347,353,367,389,401,419,431,439,449,461,

%U 467,487,499,509,523,547,563,587,599,617,631,653,661,709,727,739,761

%N Primes p(n) such that gcd(n, prime(n)+prime(n+2)) = 1.

%C This sequence and A336377 partition the set of primes.

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

%e n p(n) p(n)+p(n+2) gcd

%e 1 2 7 1

%e 2 3 10 2

%e 3 5 16 1

%e 4 7 20 4

%e 5 11 28 1

%e 6 13 32 2

%e 1 and 3 are in A336374; 2 and 4 are in A336375; 2 and 5 are in A336376; 3 and 7 are in A336377.

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

%t u = Select[Range[200], GCD[#, p[#] + p[# + 2]] == 1 &] (* A336374 *)

%t v = Select[Range[200], GCD[#, p[#] + p[# + 2]] > 1 &] (* A336375 *)

%t Prime[u] (* A336376 *)

%t Prime[v] (* A336377 *)

%Y Cf. A000040, A336366, A336374, A336375, A336377.

%K nonn

%O 1,1

%A _Clark Kimberling_, Oct 06 2020