(a*2^a)/(b*2^b)

[Don Reble]

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> The odd values seem to be of the form 2^n + odd divisor of n.

That's a helpful observation.
Let m be a divisor of n, and let
    y = 2^n,
    x = 2^n+m,
    l = n/m,
Then
    lx*2^(lx)
    --------- = x * 2^(lx-ly-n) = x
    ly*2^(ly)
So x is representable, even if the divisor is even.

Also, some (2^n + multiple of n) are representable. Let
    q = 2^(n-a)
    p = q+n
Then
    p*2^p
    ----- = p * 2^(p-q-(n-a)) = p * 2^a = 2^n + n*2^a
    q*2^q
So if a<=n, then q is integral and 2^n+n*2^a is representable.

And if n>1,
    (2n-2)*2^(2n-2)
    --------------- = 2^n
     (n-1)*2^(n-1)

Those account for the values found so far.

-- 
Don Reble  djr at nk.ca

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