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%I #12 Jan 10 2019 02:21:21
%S 0,0,1,1,1,2,1,1,1,2,2,2,2,2,1,2,1,3,2,2,2,2,1,2,2,2,2,1,2,2,2,2,2,2,
%T 2,2,3,1,1,2,1,3,2,2,2,2,3,2,1,3,2,2,3,1,1,1,2,2,3,3,2,2,2,2,3,2,3,3,
%U 1,3,2,1,2,3,2,1,2,3,2,3,2,4,2,2,2,2,2,3,3,3,1,1,1,2,2,1,2,3,2,3,3,2,1,2,3
%N Number of primes p with prime(n) == 1 (modulo 2*p).
%F a(n) = A001221(floor(A000040(n)/2)). - _Jon Maiga_, Jan 06 2019
%e n=6: prime(6)=13 and 13 mod (2*2) = 1, 13 mod (2*3) = 1, 13 mod(2*5) = 3, 13 mod (2*7) = 13, therefore a(6)=2.
%t Table[PrimeNu[Floor[Prime[n]/2]], {n, 105}] (* _Jon Maiga_, Jan 06 2019 *)
%o (PARI) a(n) = omega(prime(n)\2); \\ _Michel Marcus_, Jan 06 2019
%Y Cf. A001221, A079950, A079952.
%K nonn
%O 1,6
%A _Reinhard Zumkeller_, Jan 19 2003