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The Coprime Density: How Counting Pairs of Integers Computes 6/π², and What Golden Pi Changes¶

Almost every article in this series has taken a continuous object — a series, an integral, a product of factors — and asked where the circle constant sits inside it. This one takes something with no geometry in it at all: pairs of whole numbers. Two integers are either coprime or they are not; there is nothing to measure, no continuum to smear over, and no instrument anywhere in the question. Ask what fraction of the integer grid is coprime and a constant falls out — and that constant is the same one the integrals were converging to.

That makes it the cleanest possible test of a claim this blog keeps making: the circle constant is computed, and what is computed is 3.14159265358979323846… — a limit of exact arithmetic, not a reading off a ruler.

The question, and why a constant appears at all¶

Write \((a,b)\) for a pair of positive integers and call it coprime when \(\gcd(a,b)=1\). Ask: if you pick a pair at random, how likely is it coprime? "At random" needs care — there is no uniform probability on the positive integers — so the statement is made precise with density: let

\[P_N = \frac{\#\{(a,b) : 1 \le a,b \le N,\ \gcd(a,b)=1\}}{N^2}\]

and ask for \(\lim_{N\to\infty} P_N\). Dirichlet (1849) settled that the limit exists and equals \(6/\pi^2\); the probabilistic phrasing — "the chances are 61 to 39" — was put forward by Ernesto Cesàro in 1881 (Mathesis 1, 184) and justified in his following papers. The rigorous route is the one Dirichlet and later Mertens used; the standard reference is Hardy & Wright, An Introduction to the Theory of Numbers, Theorem 332.

Where does a circle constant come from in a counting problem about divisibility? From the primes. A prime \(p\) divides both entries of a pair with probability \(1/p^2\), so it fails to do so with probability \(1 - 1/p^2\), and divisibility by distinct primes is independent. Multiplying over all primes:

\[\prod_{p\ \text{prime}}\left(1 - \frac{1}{p^2}\right) = \frac{1}{\zeta(2)} = \frac{6}{\pi^2} = 0.607927101854026628663276779\ldots\]

That product is pure arithmetic — every factor is a rational — and its reciprocal is Euler's 1735 sum \(\zeta(2)=\sum n^{-2}=\pi^2/6\). The circle constant enters this problem through the Basel problem and nowhere else. Nothing in the counting mentions a circle.

The convergence is slow and visibly so, which is why the table below is worth printing rather than asserting:

cutoff	\(\prod_{p\le M}(1-p^{-2})\)	error vs \(6/\pi^2\)
\(p \le 10\)	0.626938775510204081633	\(+1.90\times10^{-2}\)
\(p \le 100\)	0.609033725399516660994	\(+1.11\times10^{-3}\)
\(p \le 1000\)	0.608004307306125306666	\(+7.72\times10^{-5}\)
\(p \le 100000\)	0.607927589563138868522	\(+4.88\times10^{-7}\)

Four hundred and eighty-eight parts per billion after all primes to a hundred thousand: the primes deliver the constant the way an infinite sum does, one correction at a time.

The exact count: Mertens' identity¶

The product argument is the intuitive one. The exact one is better, because it is finite arithmetic and can be evaluated. Euler's totient function \(\varphi(k)\) counts the integers in \(1..k\) that are coprime to \(k\), and summing it over \(k \le N\) counts the coprime pairs in the grid exactly:

\[\#\{(a,b): 1\le a,b\le N,\ \gcd(a,b)=1\} \;=\; 2\sum_{k=1}^{N}\varphi(k) - 1, \qquad 2\sum_{k\le N}\varphi(k)-1 = \frac{6}{\pi^2}N^2 + O(N\log N),\]

the asymptotic form being Mertens' 1874 result. Walfisz (1963) sharpened the error term to \(O(N(\log N)^{2/3}(\log\log N)^{4/3})\), and Montgomery (1987) showed the fluctuations are at least of order \(\sqrt{\log\log N}/N\) — so that error term can never simply vanish.

Computed here by totient sieve and exact integer arithmetic, with no floating point in the counts:

\(N\)	coprime pairs \(2\sum\varphi(k)-1\)	count / \(N^2\)	vs \(6/\pi^2\)	vs \(3\varphi/8\)
10	63	0.63	\(+2.21\times10^{-2}\)	\(+2.32\times10^{-2}\)
100	6 087	0.6087	\(+7.73\times10^{-4}\)	\(+1.94\times10^{-3}\)
1000	608 383	0.608383	\(+4.56\times10^{-4}\)	\(+1.62\times10^{-3}\)
10 000	60 794 971	0.60794971	\(+2.26\times10^{-5}\)	\(+1.19\times10^{-3}\)
100 000	6 079 301 507	0.6079301507	\(+3.05\times10^{-6}\)	\(+1.17\times10^{-3}\)
1 000 000	607 927 104 783	0.607927104783	\(+2.93\times10^{-9}\)	\(+1.16\times10^{-3}\)

At one million the count itself is \(607\,927\,104\,783\) and \(\sum_{k\le10^6}\varphi(k) = 303\,963\,552\,392\), whose doubling minus one returns exactly that integer. The density lands on \(0.607927104783\) against \(6/\pi^2 = 0.607927101854\): eight digits, from nothing but counting. A parenthetical honesty note: at \(N=10\) the count already leans the right way but by only five percent of the margin between the two candidate values — small grids do not decide anything, and anyone who reads a constant off a ten-by-ten grid is reading noise.

The golden relabel: the constant enters squared¶

Golden Pi is \(\hat\pi = 4/\sqrt{\varphi} = 3.144605511029693144\ldots\) with \(\varphi = (1+\sqrt5)/2\). Its square is exactly rational in the golden field:

\[\hat\pi^2 = \frac{16}{\varphi} = 8(\sqrt5 - 1) = 9.888543819998317571\ldots, \qquad \hat\pi^4 + 16\hat\pi^2 - 256 = 0.\]

Because the coprime density depends on the constant squared, the golden relabel of \(6/\pi^2\) is not a decimal accident but a closed form:

\[\frac{1}{\hat\pi^{2}} = \frac{\varphi}{16} = 0.101127124296868428\ldots, \qquad \frac{6}{\hat\pi^{2}} = \frac{3\varphi}{8} = 0.606762745781210568\ldots\]

and the gap between the two densities is the square of the gap between the two constants:

\[\frac{6/\pi^2}{6/\hat\pi^2} = \left(\frac{\hat\pi}{\pi}\right)^{\!2} = 1.001918964341353794\ldots \qquad (0.19189643\%), \qquad \frac{\hat\pi}{\pi}-1 = 0.095902230878\% .\]

So the recurring \(0.09590\%\) separation between the two labels appears here doubled to \(0.19190\%\), exactly as in Pappus's torus and the reciprocal-power family — a squared constant doubles its own error. The companion quantities follow the same rule: \(3/\pi^2 = 0.303963550927013314332\ldots\) becomes \(3/\hat\pi^2 = 3\varphi/16 = 0.303381372890605284\ldots\), so the classical coefficient \(3/\pi^2\) of the totient-sum asymptotic \(\sum_{k\le N}\varphi(k) \approx 3N^2/\pi^2\) becomes \(3\varphi/16\) under the relabel. At \(N=10^6\):

\[\sum_{k\le10^6}\varphi(k) = 303\,963\,552\,392, \qquad \frac{3N^2}{\pi^2} = 303\,963\,550\,927.013, \qquad \frac{3N^2}{\hat\pi^2} = 303\,381\,372\,890.605 .\]

The counted sum misses the analytic value by \(1\,465\); it misses the golden value by \(582\,179\,501\). Those two numbers are not in the same universe of size, and no square-root softening rescues the second one — the squared entry into the formula makes it worse, not better.

The Möbius route, and why counting beats sampling¶

The same constant has a third independent expression, by Möbius inversion: the indicator of coprimality is \(\sum_{d \mid \gcd(a,b)}\mu(d)\), which gives

\[\frac{1}{\zeta(2)} = \sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}} .\]

Truncated here in exact arithmetic: \(0.616258503401\) at \(n\le10\), \(0.608247717304\) at \(n\le100\), \(0.607931911854\) at \(n\le1000\) (\(+4.81\times10^{-6}\)), \(0.607927096761290\) at \(n\le10^5\) (\(-5.09\times10^{-9}\)) and \(0.607927102040462\) at \(n\le10^6\) (\(+1.86\times10^{-10}\) versus \(6/\pi^2\), against \(+1.16\times10^{-3}\) versus \(3\varphi/8\)). Note the alternating sign of the error: this is a genuine limit, not a monotone approach, and its fine structure is precisely what the Riemann-hypothesis connection probes.

Now the part that matters for honest accounting. A Monte Carlo estimate is the obvious way to look at this density, and it is a disaster next to the exact count. Sampling a million pairs of random integers below \(10^9\) returned \(0.607623 \pm 0.000488\) (\(-0.62\sigma\) from \(6/\pi^2\), \(+1.76\sigma\) from \(3\varphi/8\)); a hundred thousand pairs returned \(0.609180 \pm 0.001543\) (\(+0.81\sigma\) and \(+1.57\sigma\)). The gap between the labels is \(0.0011643561\), so a simple binomial estimate needs about 175 800 pairs for a one-sigma separation and 703 250 pairs for a two-sigma one — meaning that at a million samples the two labels are still less than two standard deviations apart. Sampling cannot arbitrate. Counting can, because counting has no error bar: \(607\,927\,104\,783\) integer pairs is an exact rational whose deviation from \(6/\pi^2\) is \(O(\log N/N)\) and already down to \(2.9\times10^{-9}\) at \(N=10^6\), six orders of magnitude below the \(1.16\times10^{-3}\) golden offset. The integer grid decides this one, and it decides it for the analytic constant.

What the constant is not doing here¶

Two of the three cousins of this density are entirely π-free, and printing them is part of keeping the site honest about what is really carrying the constant.

Setwise coprimality of \(k\) integers is \(1/\zeta(k)\): \(1/\zeta(3) = 0.831907372580707469\ldots\) and \(1/\zeta(4) = 0.923938402921590167\ldots\). For \(k\ge3\) the answer has no circle constant in it at all — it is \(\zeta(3)\), \(\zeta(4)\), irreducibly.
Pairwise coprimality — a stronger condition, that every pair in the set is coprime — has density \(\prod_p\left(1 - 3p^{-2} + 2p^{-3}\right) = 0.2867477554715617858\ldots\) for three integers. That product is arithmetic over the primes; no \(\pi\) appears in it anywhere, under either label.
Euclid's orchard: coprime pairs are exactly the lattice points visible from the origin, the ones no nearer lattice point shadows. That picture is geometry, and it too is constant-free — visibility is a property of \(\gcd\), not of any turn.

So the circle constant is not "everywhere in number theory". It is in exactly one place here: the infinite product over the primes, whose reciprocal is \(\zeta(2)\), and whose evaluation is Euler's sum. That is a computation — an exact limit — and the claim that it is a measurement would be false on its face, since nothing in this article has units.

The honest boundary¶

The ledger, stated plainly:

statement	status
Coprime pairs have density \(6/\pi^2\) (Dirichlet 1849; Cesàro 1881)	exact, and the count confirms it to eight digits by \(N=10^6\)
The density is \(6/\hat\pi^2 = 3\varphi/8\) under Golden Pi	exact in the golden field — \(\hat\pi^2 = 16/\varphi\) — and \(0.19190\%\) away from the counted value
\(3\varphi/16\) replaces \(3/\pi^2\) in \(\sum\varphi(k)\)	exact algebra; the counted sum at \(10^6\) misses it by \(5.8\times10^8\)
Pairwise-coprime and \(k\ge3\) densities	π-free — nothing for either label to change
Monte Carlo can decide between the labels	false: \(\sim1.8\sigma\) at a million pairs
Exact counting can decide	true in principle, and it does so at \(N\ge10^4\)

What this example does show the golden case is the algebraic exactness that the analytic constant does not have: \(\hat\pi\) is a root of \(x^4 + 16x^2 - 256\) and is constructible from \(\varphi\) by square roots, so the golden density \(3\varphi/8\) is a closed-form algebraic number rather than a transcendental limit. That is real, and it is credited. What it does not show is that the number of coprime pairs in a grid of integers is \(0.606762745781\ldots\): it is \(0.607927104783\) at a million, and the difference is arithmetic, not experimental. The circle constant was computed here, on both sides of the ledger — never measured, by either side.