Let b(0)=1; b(1)=1; b(n+2)=(e^g+1/e^g)*b(n+1)-b(n). a(n)=floor(b(n)).
A093608
Let b(0)=1; b(1)=1; b(n+2)=(e^g+1/e^g)*b(n+1)-b(n). a(n)=floor(b(n)).
Terms
- a(0) =1a(1) =1a(2) =2a(3) =3a(4) =6a(5) =11a(6) =20a(7) =36a(8) =64a(9) =115a(10) =205a(11) =366a(12) =652a(13) =1162a(14) =2070a(15) =3687a(16) =6567a(17) =11696a(18) =20832a(19) =37103a(20) =66084a(21) =117701a(22) =209635a(23) =373375a(24) =665008a(25) =1184428a(26) =2109552a(27) =3757265a(28) =6691962a(29) =11918868
External references
- oeis: A093608