The following is a proof that A010061 [binary self numbers] does not contain any consecutive values. The proof may be adapted for b-ary self numbers.


We will prove that the complement of A010061 has a maximal difference of two. A pair of consecutive elements would lead to a difference of at least three.
Specifically, we prove for all n that if n is NOT a self number, then either n+1 or n+2 is not a self number.

For ease of use let d(n) be the sum of binary digits of n.

We will proceed by induction.

Say the theorem is true for all n < N, for some N.

Suppose N is not a self number, so N = t + d(t) for some t < N.


Now let 0 <= k <= t < N be the largest integer with t = 2^k - 1 mod 2^k.

This corresponds to the lowest zero-bit of N.
Then write t = 2^(k+1) X + 2^k - 1, where X >= 0.

Now since k = d(2^k - 1) <= 2^k - 1 <= t < N, either k or k-1 is not a self number.

Thus, there exists j < k < 2^k$ with j+d(j) equal to either k-1 or k.

Finally write r = 2^(k+1) X + 2^k + j.

Doing some calculations, N = t + d(t) = 2^(k+1) X + 2^k - 1 + d(X) + k.
Also r + d(r) = 2^(k+1) X + 2^k + j + d(X) + 1 + d(j).
We see that r + d(r) = N - k + 2 + j + d(j).
If j + d(j) = k, then r + d(r) = N + 2.
If j + d(j) = k - 1, then r + d(r) = N + 1.
Thus, either N + 1 is not a self number or N + 2 is not a self number, so the theorem is true for N.

- Griffin N. Macris, May 31 2020