Re: mu(Large)
Would taking several zeta zeros improve things? Maybe with weighted voting?
Why doesn't this give the same estimates for mu(Large) and mu(Large+1) and ... ?
One might also try finding primes P that don't divide Large and
guestimating mu(P*Large); this would reflect on the likely value of mu(Large).
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sums of squares: for Dan's problem:
Two squares, N is x2+y2 iff all the poison divisors of N (those P=4K+3)
occur to an even power. 9 is ok as an exact power divisor (quotient not
a multiple of 3), but 3 and 27 are out.
As N gets large, 0% are the sum of two squares;
if a number is the sum of two squares, it's probably the sum many ways.
1105 = 5.13.17 = four ways, or 32 ways with +- signs and swapping.
Of course the total number of sums of squares < N (counting multiplicity)
is roughly pi*N (or pi/4 or pi/8, depending on sign & ordering convention).
So the average number of representations of a number is pi, but the
representations concentrate in a few values.
Three squares, N is x2+y2+z2, if, after removing all possible divisors
of 4, the leftover is not 8K+7.
The typical number of representations is roughly sqrtN, but I think
it can wander up & down by a slowly growing factor, logN or loglogN.
When 4|N, x,y,z must be even, so 4^K has only one "unique" rep, and
4^K N and N have the same number of reps.
Four squares, all N are representable, the number of reps is a simple
multiple of the sum of the divisors of N, with the multiplier depending
roughly on N mod 8. Divsum(N)/N >=1, and is less than a slowly growing
function of N.
Overall, these functions are fairly irregular, so I think Dan's
program will work better in in higher dimensions. Hardy & Wright sez
formulas exist for 5,6, and 8 squares. Conway & Sloane will know more,
if they are telling.
Rich rcs(a)cs.arizona.edu
