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). ---- 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@cs.arizona.edu