%I #32 Dec 15 2023 18:11:01
%S 7,1,2,3,2,5,3,7,2,3,5,1,3,1,7,5,2,1,3,1,5,7,2,1,3,5,2,3,7,1,5,1,2,3,
%T 2,7,3,1,2,3,5,1,7,1,2,5,2,1,3,7,5,3,2,1,3,5,7,3,2,1,5,1,2,7,2,5,3,1,
%U 2,3,7,1,3,1,2,5,2,7,3,1,5,3,2,1,7,5,2
%N The integers sieved by 7, 5, 3, and 2.
%C A number is tested for the following in order - the first test passed determines a(n):
%C Is n mod 7 == 1? If so, write 7.
%C Is n mod 5 == 1? If so, write 5.
%C Is n mod 3 == 1? If so, write 3.
%C Is n mod 2 == 1? If so, write 2.
%C Write 1.
%C Period 210. - _Charles R Greathouse IV_, Sep 22 2012
%C An example of an inverted sieve. Usually we would sieve 2, 3, 5, 7 which gives 2, 1, 2, 3, 2, 5, 2, 7, 2, 3, 2, 1. - _Jon Perry_, Sep 24 2012
%e 4 is not 1 mod 7 or 1 mod 5 but is 1 mod 3, so a(4) = 3.
%t Table[If[Mod[n, 7] == 1, 7, If[Mod[n, 5] == 1, 5, If[Mod[n, 3] == 1, 3, If[Mod[n, 2] == 1, 2, 1]]]], {n, 100}] (* _T. D. Noe_, Sep 25 2012 *)
%t Table[Which[Mod[n,7]==1,7,Mod[n,5]==1,5,Mod[n,3]==1,3,Mod[n,2]==1,2,True,1],{n,90}] (* _Harvey P. Dale_, Mar 04 2016 *)
%o (JavaScript) for (i=1;i<90;i++)
%o if (i%7==1) document.write("7,");
%o else if (i%5==1) document.write("5,");
%o else if (i%3==1) document.write("3,");
%o else if (i%2==1) document.write("2,");
%o else document.write("1,");
%o (C++)
%o template <typename T> std::string PrintSequnce_A216983(const T &max)
%o {
%o std::string strSeq;
%o for(T i = 1; i < max; ++i)
%o {
%o if (i%7==1) strSeq+="7";
%o else if (i%5==1) strSeq+="5";
%o else if (i%3==1) strSeq+="3";
%o else if (i%2==1) strSeq+="2";
%o else strSeq+="1";
%o if(i<max-1)
%o strSeq+=", ";
%o }
%o return strSeq;
%o } // _Martin Ettl_, Oct 08 2012
%K nonn,easy
%O 1,1
%A _Jon Perry_, Sep 21 2012