a(1) = 1; a(n+1) = (product{k|n} a(k)) (sum{j|n} 1/a(j)), where both the product and sum are over the positive divisors of n.
A068342
a(1) = 1; a(n+1) = (product{k|n} a(k)) (sum{j|n} 1/a(j)), where both the product and sum are over the positive divisors of n.
Terms
- a(0) =1a(1) =1a(2) =2a(3) =3a(4) =7a(5) =8a(6) =42a(7) =43a(8) =304a(9) =914a(10) =13717a(11) =13718a(12) =1948004a(13) =1948005a(14) =165580467a(15) =3808350755
External references
- oeis: A068342