Congruent Products Under XOR; Fibbinary Numbers

[Antti Karttunen]

franktaw at netscape.net wrote:

> I think you have ahold of the wrong end of the stick here.
>  
> if 2*i XOR 5*i = 3*i, then 2*i XOR 3*i = 5*i.

> What I'm seeing here is cases where a*i and b*i have no common bits, 
> so a*i XOR b*i = (a+b)*i.

Actually, this is a very promising lead, as the pairs (a,b):
(3,7), (7,11), (13,21), (15,23),
(for which there exists non-zero solutions to the "congruence"
a*n = B X n (where X is the carryless binary multiplication A048720))

allow the following decompositions, where the right hand side is
decomposed to two integers with disjoint bit patterns, so that
the smaller element plus the left hand side is equal to the larger element:

3,7 -> 3, 2 XOR 5 -> 5 = 2+3
7,11 -> 7, 2 XOR 9 -> 9 = 2+7
13,21 -> 13, 4 XOR 17 -> 17 = 4+13
15,23 -> 15, 4 XOR 19 -> 19 = 4+15

However, the sequence generated with say a condition
"i such that 2*i XOR 7*i = 9*i" doesn't necessarily generate
exactly the same set of integers as in

 >%S A115770 
0,7,14,15,28,30,31,56,60,62,63,112,120,124,126,127,224,240,248,252,
 >%T A115770 
254,255,448,455,480,496,504,508,510,511,896,903,910,911,960,967,992,
 >%U A115770 
1008,1016,1020,1022,1023,1792,1799,1806,1807,1820,1822,1823,1920
 >%N A115770 Integers i such that 7*i = A048720bi(11,i).

but could be a superset of it (as A115767 is a superset of A048719). 
Have to compute (and maybe
to use my brains also a bit, for a change). Will return.


Terveisin,

Antti