bounces

[hv at crypt.org]

The following comment is incorrect and should be removed or corrected:

%C A064541 There is a unique number with m digits for each m >= 2. If
a(n) has m digits, then               a(n+1) is the number given by
the final m+1 digits of 2^a(n). - Eric Rains and njas, Oct 08, 2001

In particular, there is NO 15-digit term of A064541.
The 14-digit term is A064541(13) = 75353432948736, and 2^A064541(13)
ends with ...5075353432948736.
Therefore, it does contribute a 15-digit term to A064541, the next
term is 16-digit number:
A064541(14) = 5075353432948736.

Correspondingly, the following formula is incorrect either:

%F A121319 a(n) = A064541(n-1) for n>1. - Alexander Adamchuk
(alex(AT)kolmogorov.com), Jan 28 2007

In particular, A121319 has the following terms:
A121319(14) = 75353432948736
A121319(15) = 5075353432948736
A121319(16) = 5075353432948736
i.e., A121319 is repeating some values two or more times while A064541
does not repeat any values (as follows from its definition: "%N
A064541 Values of n such that 2^n ends in n.").

Max