Conjectures 111-113 from "100 Conjectures from the OEIS"

[David Wilson]

> I would like to see your method, as mine was so clumsy.

A number n is divisible by b+1 iff the alternating sum of its digits is 
divisible by b+1.  This is the basis for "casting out elevens", a base-10 
test for divisibility by 11.  The base 2 case says that a number is 
divisible by 3 iff the alternating sum of its base-2 digits is divisible by 
3, that is to say, n is divisible by 3 iff A065359(n) is divisible by 3.

Also, it is easy to show that a base 2 number has an even (odd) number of 
1's in its base 2 representation iff the alternating sum of its base 2 
digits is even (odd). That is, n is evil (odious) iff A065359(n) is even 
(odd).

Together, these observations allow us to conclude that n is an evil (odious) 
multiple of 3 iff A065359(n) is an even (odd) multiple of 3.

So if A065359(n) = 3 or -3, then n is an odious multiple of 3, and if 
A065359(n) = 0, then n is an evil multiple of 3.  That is to say, 
conjectures 111 and 112 are true.

Since A065359((4^n-1)/3) = n and A065359(2*(4^n-1)/3) = -n, the range of 
A065359 is Z. Specifically, there will be some n for which A065359(n) = 9, 
this n must be an odious multiple of 3. This n is an odious multiple of 3 
for which A065359(n) is neither 3 nor -3.  Likewise, there exists n with 
A065359(n) = 6, this n is an evil multiple of 3 for which A065359(n) is not 
0.  This means that Ralf's conjectures on A065359 are incorrect (because the 
implication does not apply in the backward direction).