A095389 a(n) is the number of residues from reduced residue system, R, modulo 210 such that both R and R+2 are primes, i.e., both 210n+r and 210n+r+2 are primes at fixed n.

13, 7, 6, 5, 5, 4, 6, 5, 5, 6, 6, 2, 6, 2, 3, 6, 7, 3, 4, 6, 6, 4, 5, 4, 2, 3, 6, 4, 1, 4, 2, 5, 5, 3, 4, 4, 2, 2, 2, 4, 3, 2, 5, 2, 5, 2, 4, 4, 3, 5, 2, 2, 4, 2, 3, 2, 4, 4, 3, 1, 1, 4, 1, 2, 0, 6, 5, 2, 3, 4, 1, 0, 4, 1, 5, 1, 4, 3, 1, 3, 3, 3, 3, 3, 5, 7, 3, 2, 2, 0, 3, 3, 4, 2, 3, 4, 2, 4, 4, 3, 4, 2, 6, 3, 1

Offset: 0

Author

Labos Elemer and Enoch Haga, Jun 15 2004

Keywords

• nonn

Comments

Since arbitrarily large prime gaps occur, several consecutive zeros may arise in the sequence.

Examples

			n=0: only 13+2=15 integers correspond to the condition: {11,17,29,41,59,71,101,107,137,149,179,191,197}, so a[0]=13; see A078859.
n=11: only 2 twins were found, {2339,2341} and {2381,2383} corresponding to residue pairs {29,31} and {71,73}.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A001359, A078859.

Programs

Mathematica

{k =0, ta=Table[0, {100}]}; Do[{m=0};Do[s=210k+r;s1=210k+r+2; If[PrimeQ[s]&&PrimeQ[s+2], m=m+1], {r, 1, 210}];ta[[k]]=m, {k, 1, 100}];ta
(* Second program: *)
With[{P = Product[Prime@ i, {i, 4}]}, Function[R, Array[Count[R + P #, k_ /; Times @@ Boole@ PrimeQ@ {k, k + 2} == 1] &, 105, 0]]@ Select[Partition[Select[Range[P + 1], CoprimeQ[#, P] &], 2, 1], Differences@ # == {2} &][[All, 1]]] (* Michael De Vlieger, May 15 2017 *)