Least k such that 1/tau(k) + 1/tau(k+1) + 1/tau(k+2) + ... + 1/tau(k+n) is equal to 1 (where tau(k)=A000005(k) is the number of divisors of k).

A073545

Least k such that 1/tau(k) + 1/tau(k+1) + 1/tau(k+2) + ... + 1/tau(k+n) is equal to 1 (where tau(k)=A000005(k) is the number of divisors of k).

Terms

    a(0) =1a(1) =2a(2) =6a(3) =25a(4) =54a(5) =243a(6) =1204a(7) =3549a(8) =19544a(9) =81829a(10) =104663a(11) =663490a(12) =743764a(13) =7925355a(14) =15376922a(15) =39462786a(16) =201432540a(17) =1187707803a(18) =3034296474a(19) =8657654859a(20) =48511905236a(21) =154669032693a(22) =123533546264

External references