Is it a Proth number? A Proth number, named after François Proth, is a number that can be expressed as
$$N = k \times 2^n + 1$$
Where \$k\$ is an odd positive integer and \$n\$ is a positive integer such that \$2^n &gt; k\$. Let's use a more concrete example. Take 3. 3 is a Proth number because it can be written as
$$ 3 = (1 \times 2^1) + 1 $$
and \$2^1 &gt; 1\$ is satisfied. 5 Is also a Proth number because it can be written as
$$ 5 = (1 \times 2^2) + 1 $$
and \$2^2 &gt; 1\$ is satisfied. However, 7 is not a Proth number because the only way to write it in the form \$N = k \times 2^n + 1\$ is
$$ 7 = (3 \times 2^1) + 1 $$
and \$2^1 &gt; 3\$ is not satisfied.
Your challenge is fairly simple: you must write a program or function that, given a positive integer, determines if it is a Proth number or not. You may take input in any reasonable format, and should output a truthy value if it is a Proth number and a falsy value if it is not. If your language has any &quot;Proth-number detecting&quot; functions, you may use them.
Test IO
Here are the first 46 Proth numbers up to 1000. (A080075)
3, 5, 9, 13, 17, 25, 33, 41, 49, 57, 65, 81, 97, 113, 129, 145, 161, 177, 193, 209, 225, 241, 257, 289, 321, 353, 385, 417, 449, 481, 513, 545, 577, 609, 641, 673, 705, 737, 769, 801, 833, 865, 897, 929, 961, 993

Every other valid input should give a falsy value.
As usual, this is code-golf, so standard loopholes apply, and the shortest answer in bytes wins!

Number theory fun-fact side-note:
The largest known prime that is not a Mersenne Prime is \$19249 \times 2^{13018586} + 1\$, which just so happens to also be a Proth number!
Jelly, 5 bytes
’&amp;C²&gt;

Try it online! or verify all test cases.
Let \$\newcommand{\&amp;}[]{\text{ &amp; }}j\$ be a strictly positive integer. \$j + 1\$ toggles all trailing set bits of \$j\$ and the adjacent unset bit. For example, \$10011_{2} + 1 = 10100_{2}\$.
Since \${\sim}j = -(j + 1) = -j - 1\$, \$-j = {\sim}j + 1\$, so \$-n\$ applies the above to the bitwise NOT of \$j\$ (which toggles all bits), thus toggling all bits before the last \$1\$.
By taking \$j \&amp; (-j)\$ (the bitwise AND of \$j\$ and \$-j\$), all bits before and after the last set bit are nullified (since unequal in \$j\$ and \$-j\$), thus yielding the highest power of \$2\$ that divides \$j\$ evenly.
For input \$N\$, we want to apply the above to \$N - 1\$ to find \$2^{n}\$, the highest power of \$2\$ that divides \$N - 1\$. If \$m = N - 1\$, \$-m = -(N - 1) = 1 - N\$, so \$(N - 1) \&amp; (1 - N)\$ yields \$2^{n}\$.
All that's left to test is if \$2^{n} &gt; k\$. If \$k &gt; 0\$, this is true if and only if \$(2^{n})^{2} &gt; k2^{n}\$, which is true itself if and only if \$(2^{n})^{2} \ge k2^{n} + 1 = N\$.
Finally, if \$(2^{n})^{2} = N = k2^{n} + 1\$, \$2^{n}\$ must be odd (\$1\$) so the parities of both sides can match, implying that \$k = 0\$ and \$N = 1\$. In this case \$(N - 1) \&amp; (1 - N) = 0 \&amp; 0 = 0\$ and \$((N - 1) \&amp; (1 - N))^{2} = 0 &lt; 1 = N\$.
Therefore, \$((N - 1) \&amp; (1 - N))^{2} &gt; N\$ is true if and only if \$N\$ is a Proth number.
How it works
’&amp;C²&gt;  Main link. Argument: N

’      Decrement; yield N - 1.
  C    Complement; yield 1 - N.
 &amp;     Take the bitwise AND of both results.
   ²   Square the bitwise AND.
    &gt;  Compare the square to N.