a(1)=1; thereafter a(n+1) = a(n) + Product_{b=2..n+1} b^(1+floor(log_b(a(i)))).

A319955

a(1)=1; thereafter a(n+1) = a(n) + Product_{b=2..n+1} b^(1+floor(log_b(a(i)))).

Terms

    a(0) =1a(1) =3a(2) =39a(3) =331815

External references