a(0) = 1, then (for n>0) a(n) = floor[(e + 1/e)*a(n-1) - a(n-2)].
A085560
a(0) = 1, then (for n>0) a(n) = floor[(e + 1/e)*a(n-1) - a(n-2)].
Terms
- a(0) =1a(1) =3a(2) =8a(3) =21a(4) =56a(5) =151a(6) =410a(7) =1114a(8) =3027a(9) =8227a(10) =22362a(11) =60785a(12) =165230a(13) =449141a(14) =1220891a(15) =3318725a(16) =9021229a(17) =24522242a(18) =66658364a(19) =181196219a(20) =492542389a(21) =1338869025a(22) =3639423341a(23) =9892978333a(24) =26891903231a(25) =73099771885a(26) =198705781579
External references
- oeis: A085560