updated 3 sequences

[N. J. A. Sloane]

Numbers n such that the product of the digits of n
is equal to the totient of the reverse of n

1, 42, 62, 78, 495, 666, 861, 883, 6876, 8795, 8862, 8887, 8965, 8991, 69786

Indeed, phi(5978) = 2520 = 8*7*9*5.

giovanni





There was some discussion here yesterday on this topic.

My colleague David Applegate (not currently a memeber
of this list) comments that the correct generalization is the following:


The appropriate generalization of floor(2/(2^(1/n)-1)) =? ceil(2n/log(2))
is a/(b^(1/n)-1) approx an/log(b) - a/2.

When a=2, the a/2 becomes 1, and can be hidden in floor() = ceil() - 1.
Doing floor=ceil with a != 2 is just wrong - almost every n will
work.  

(Not that either of feel that these generalizations should be added
to the OEIS!)

Neil