- Let $a(n)$ be A260711 without initial $0$ (i.e., numbers of the form $x^2 - y^2$ with $x > y$ where $x$ and $y$ are odd, $x + y$ is a power of $2$).
The sequence begins with
$$ 8, 16, 32, 48, 64, 96, 128, 160, 192, 224, 256, 320, 384, 448, 512 $$
- Let
$$ \ell(n) = \left\lfloor\log_2 n\right\rfloor $$
Let $\operatorname{tr(n)}$ be A007814 (i.e., number of trailing zeros in the binary expansion of $n$).
The sequence begins with
$$ 1, 1, 1, 3, 1, 3, 1, 5, 3, 7, 1, 5, 3, 7, 1 $$
The sequence begins with
$$ 1, 2, 3, 2, 4, 3, 5, 3, 4, 3, 6, 4, 5, 4, 7 $$
- Let $c(n)$ be A162751 (i.e., write down in binary the $n$-th positive (odd) integer that is a palindrome in base $2$. Take only the leftmost half of the digits (including the middle digit if there are an odd number of digits). $c(n)$ is the decimal equivalent of the result).
The sequence begins with
$$ 1, 1, 2, 3, 2, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8 $$
There are no formulas for $b_1(n)$ and $b_2(n)$ in the OEIS, so after a little inspection I conjecture that
$$ c(n) = 2^{\ell(n+1)-1} + (n+1)\operatorname{mod}2^{\ell(n+1)-1}, \\ b_1(n) = \frac{c(n)}{2^{\operatorname{tr}(c(n))}}, \\ b_2(n) = \ell(n+1) + \operatorname{tr}(c(n)) + [(n+1) \geqslant 3\cdot2^{\ell(n+1)-1}] $$
Here square bracket denotes Iverson bracket.
I also conjecture that
$$ a(n) = c(n)2^{\ell\left(\left\lfloor\frac{2(n+1)}{3}\right\rfloor\right)+3} $$
and the smallest (and possibly unique) solution for $a(n)$ is
$$ x_n = 2^{b_2(n)} + b_1(n), \\ y_n = 2^{b_2(n)} - b_1(n) $$
Note that it looks like that $\left\lfloor\frac{x_n}{2}\right\rfloor$ is A120242.
Here is the PARI/GP program to check it numerically:
a(n) = n++; my(L = logint(n, 2), A = 2^(L-1)); (A + n%A)*2^(logint(2*n\3, 2) + 3)
b1(n) = n++; my(L = logint(n, 2), A = 2^(L-1)); A += n % A; A /= 2^valuation(A, 2)
b2(n) = n++; my(L = logint(n, 2), A = 2^(L-1)); valuation(A + n % A, 2) + L + bittest(n, L-1)
test(n) = my(A = 2^b2(n), B = b1(n)); a(n) == (A + B)^2 - (A - B)^2
Is there a way to prove it?
