... but I don't think a complete proof.
I made a list of the first 10,000 Padovans and the first 10,000 numbers which are
one less than a power of two.
The only qualifying Padovans were [1, 3, 7]
But this is not a proof.
I note that in order to get 3: 1 and 2 have been added.
In order to get 7: 3 and 4 have been added.
In these cases there were consecutive Padovans which were:
1 less than a power of 2 followed by a power of 2.
In general, if binary a+b=c (a<b) and c is all 1's, then a must be the 1's complement of b.
Further, a must be one less binary digit than b (or you could think of it as though a has one leading zero).
As the Padovan numbers grow larger, looking at the ratio between near-by elements:
the ratio of the (i-3) term i-th term is about .43
the ratio of the (i-2) term i-th term is about .57
the ratio of the (i-1) term i-th term is about .75
The first two of these ratios are close to 3/7 and 4/7 so it is not surprising that there was a 7 formed by adding 3 and 4.
But with larger numbers, there cannot be any similar pairs (2^n) - 1, 2^n, X,(2^(n+1))-1.
But this still does not prove that there could be an adjacent pair of Padovans like:
(2^n) - 1 - k, 2^n + k
So I don't think this is a full proof.
Answer with attempt at proof
