Generate all combinations for pair of bits set to 1?

Question

I'm trying to generate all possible combinations for pair of 1's within given bit width.

Let's say the bit width is 6, i.e. number 32. This is what I would like to generate:

000000
000011
000110
001100
001111
011000
011011
011110
110000
110011
110110
111100
111111

If I have variables:

var a = 1,
    b = 2;
    num = a | b;

and create a loop that I'll loop over width - 1 times, and where I shift both a << 1 and b << 1, I'll get all combinations for one pair. After that, I'm pretty much stuck.

Could someone , please, provide some help.

Update: working example
Based on Barmar's mathematical approach, this is what I managed to implement

var arr = [],
    arrBits = [];

function getCombs(pairs, startIdx) {
    var i, j, val = 0, tmpVal, idx;

    if (startIdx + 2 < pairs) {
        startIdx = arr.length - 1;
        pairs -= 1;
    }

    if (pairs < 2) {
        return;
    }

    for (i = 0; i < pairs-1; i++) {
        idx = startIdx - (i * 2);
        val += arr[idx];

    }

    for (j = 0; j < idx - 1; j++) {
        arrBits.push((val + arr[j]).toString(2));
    }

    getCombs(pairs, startIdx-1);
}

(function initArr(bits) {
    var i, val, pairs, startIdx;

    for (i = 1; i < bits; i++) {
        val = i == 1 ? 3 : val * 2;
        arr.push(val);
        arrBits.push(val.toString(2));
    }

    pairs = Math.floor(bits / 2);
    startIdx = arr.length - 1;

    getCombs(pairs, startIdx);
    console.log(arrBits);

}(9));

Working example on JSFiddle
http://jsfiddle.net/zywc5/

Answer 1

The numbers with exactly one pair of 1's are the sequence 3, 6, 12, 24, 48, ...; they start with 3 and just double each time.

The numbers with two pairs of 1's are 12+3, 24+3, 24+6, 48+3, 48+6, 48+12, ...; these are the above sequence starting at 12 + the original sequence up to n/4.

The numbers with three pairs of 1's are 48+12+3, 96+12+3, 96+24+3, 96+24+6, ...

The relationship between each of these suggests a recursive algorithm making use of the original doubling sequence. I don't have time right now to write it, but I think this should get you going.