Is it known that the primes on the Ulam prime spiral
distribute themselves equally
in sectors around the origin?
To be specific, say the quadrants?
(Each quadrant is closed on one axis and open on the other.)
For example, in the $50 \times 50$ spiral below,
I count the number of primes in the four quadrants to be $(103,96,88,86)$ ($\sum=373$), leading to ratios $(0.276,0.257,0.236,0.231)$:
Continuing to $1500 \times 1500$, which includes $166169$ primes,
I count $(41561,41528,41685,41395)$ leading to
$(0.2501,0.2499,0.2509,0.2491)$:
I earlier asked
this question at MSE, where
user Fred Daniel Kline
reported carrying out the computation to include 1.7 billion primes,
when the quadrant ratios fall within $\frac{1}{4} \pm 10^{-6}$.
Empirically there is a convergence to $\frac{1}{4}$,
but I wonder if this has been proven?
Thanks!
Best Answer
I haven't looked at the papers in detail, but apparently this paper of Wolke and also this paper of Stux establish that the square roots $p^{1/2}$ of primes are uniformly distributed modulo 1. As the fractional part of $p^{1/2}$ asymptotically determines the argument along the Ulam spiral, this should give the desired uniform distribution. (Such results also be obtainable from the known versions of the prime number theorem in short intervals such as $[x,x+x^{1/2}]$ on the average, which in turn should follow from the known zero density theorems for the zeta function (as discussed, for instance, in the Iwaniec-Kowalski book).)