N = k * 2^n + 1

(1 * 2^1) + 1

(1 * 2^2) + 1

(3 * 2^1) + 1

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

’&C²>

’&C²>  Main link. Argument: N

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

1<#<4^IntegerExponent[#-1,2]&

Reap[Do[If[f[i],Sow[i]],{i,1,1000}]][[2,1]]

{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}

BitAnd[#-1,1-#]^2>#&

<©Ó¬oD®s/›

[2, 2, 2, 2, 3, 5]

[2, 2, 2, 2]

[3, 5]

qtYF1)EW<

q    % Input x implicitly. Subtract 1
t    % Duplicate
YF   % Exponents of prime factorization of x-1
1)   % First entry: exponent of 2. Errors for x equal to 1 or 2
E    % Duplicate
W    % 2 raised to that. This is y squared
<    % Is x-1 less than y squared? Implicitly display

f x=length [x|k<-[1,3..x],n<-[1..x],k*2^n+1==x,2^n>k]>0

f x=or[k*2^n+1==x|k<-[1,3..x],n<-[1..x],2^n>k]

({}[()])(((<>()))){{}([(((({}<(({}){})>){}){})<>[({})(())])](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}}(<{}{}>)<>{({}[()])<>(({}()[({})])){{}(<({}({}))>)}{}<>}{}<>({}<>)

({}[()])      #Subtract one from input
(((<>())))    #Put three ones on the other stack
{
 {}           #Pop the crap off the top
 ([(
  ((({}<(({}){})>){}){}) #Multiply the top by four and the bottom by two
  <>[({})(())])](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{} #Check if the power of four is greater than N-1
}
(<{}{}>) #Remove the power of 4
<>{({}[()])<>(({}()[({})])){{}(<({}({}))>)}{}<>}{}<>({}<{}><>) #Modulo N-1 by the power of two

({}[()])(((<>()))){{}([(((({}<(({}){})>){}){})<>[({})(())])](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}}(<{}{}>)<>{({}[()])<>(({}()[({})])){{}(<({}({}))>)}{}<>}{}<>({}<{}><>)

>N>0,2:N^P:K*+?,P>K:2%1,N:K=

>N>0,2:N^P:K*+?,P>K:2%1,N:K=

>N>0                           input > N > 0
     2:N^P                     2^N = P
         P:K*+?                P*K+1 = input
                P>K            P > K
                  K:2%1        K%2 = 1
                        N:K=   [N:K] has a solution

!x=~-x&-~-x>x^.5

x=scan()-1;n=0;while(!x%%2){x=x/2;n=n+1};2^(2*n)>x

%:<<:AND-.

f =: %:<<:AND-.
   f 16
0
   f 17
1
   (#~f"0) >: i. 100  NB. Filter the numbers [1, 100]
3 5 9 13 17 25 33 41 49 57 65 81 97

%:<<:AND-.  Input: n
        -.  Complement. Compute 1-n
   <:       Decrement. Compute n-1
     AND    Bitwise-and between 1-n and n-1
%:          Square root of n
  <         Compare sqrt(n) < ((1-n) & (n-1))

boolean g(int p){return p--<(p&-p)*(p&-p);}

boolean f(int p){return(p-1&(1-p))>Math.sqrt(p);}

boolean g(int p){return Math.pow(p-1&(1-p),2)>p;}

double g(int p){return Math.pow(p-1&(1-p),2)-p;}

((p - 1 & (1 - p))^2) > p;

qi_(_W*&2#<

IsProth:=proc(X)local n:=0;local x:=X-1;while x mod 2<>1 do x:=x/2;n:=n+1;end do;is(2^n>x);end proc:

IsProth := proc( X )
    local n := 0;
    local x := X - 1;
    while x mod 2 <> 1 do
        x := x / 2;
        n := n + 1;
    end do;
    is( 2^n > x );
end proc:

x=>x--<(-x&x)**2

d+
$*
+`(1+)1
$+0
01
1
+`.10(0*1)$
1$1
^10*1$

d+
$*

+`(1+)1
$+0
01
1

+`.10(0*1)$
1$1

^10*1$