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Primes p(n) such that gcd(n, prime(n)+prime(n+1)) = 1.
4

%I #4 Oct 04 2020 23:34:20

%S 2,11,17,23,31,41,59,67,73,83,97,109,127,157,167,179,191,211,227,233,

%T 241,277,283,331,353,367,389,401,431,439,461,467,499,509,523,547,563,

%U 587,599,607,617,631,653,661,677,691,709,727,739,751,773,797,829,859

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

%C This sequence and A336369 partition the set of primes.

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

%e n p(n) p(n)+p(n+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 1 and 5 are in A336366; 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 *)

%Y Cf. A000040, A336366, A336367, A336369.

%K nonn

%O 1,1

%A _Clark Kimberling_, Oct 04 2020