ATA A thinking ape math challenge

//First I created a simulation to verify my understanding of this problem

function f(n){
    var set = [];
    for (var i=0;i<n;i++){//initialize circle
        set.push(i+1);
    }
    //starting by the second
    //until there is one left
    //remove current position
    //so by moving 1 I skip 1
    for (var i = 1; (set.length>1);i = (i+1)%set.length){
        set.splice(i, 1);
        console.log(set);
    }
    return set[0];
}
/*
By doing this I had noticed a pattern:

every base 2 number returns 1 while the other are the nth odd number after last base 2 result and next.

nth odd formula: odd(n) = n*2+1

for (var i=1;i<65;i++) console.log("f("+i + ")=" + f(i));
f(1)=1
f(2)=1      2^1
f(3)=3      1º odd(1) = 1*2+1 = 3
f(4)=1      2^2
f(5)=3      1º odd(1) = 1*2+1 = 3
f(6)=5      2º odd(2) = 2*2+1 = 5
f(7)=7      3º odd(3) = 3*2+1 = 7
f(8)=1      2^3
f(9)=3      1º odd(1) = 1*2+1 = 3
f(10)=5     2º odd(2) = 2*2+1 = 5
f(11)=7     3º odd(3) = 3*2+1 = 7
f(12)=9     4º odd(4) = 4*2+1 = 9
f(13)=11    5º odd(5) = 5*2+1 = 11
f(14)=13    6º odd(6) = 6*2+1 = 13
f(15)=15    7º odd(7) = 7*2+1 = 15
f(16)=1     2^4

*/

//therefore, there is always an a that is 2^a<=n<2^(a+1)
//since, when n is equal 2^a, the result is odd(0)=1, we can use the remaining of a division by 2^a
//therefore, it is the final formula:
// f(n) = odd(n mod 2^a)
//which is f(n) = (n mod 2^a)*2+1 where 2^a<=n<2^(a+1)
// here is my code implementation
function f2(n){
    var a=0;
    while(1<<a<=n) a++;//find 'a'
    a--;
    return (n%(1<<a))*2+1;// (n mod 2^a)*2+1
}

//my unit test to validate my formula

function validate (f, f2, n){
    var i=3;
    var flg = true
    while (i++<n && flg){
        flg = flg && f(i)==f2(i)
    }
    return flg;
}

f(n) = (n mod 2^a)*2+1 where 2^a<=n<2^(a+1)