Pi Normality

Pi Normality

Definitions and mathematical statement

For an integer base $b\ge2$, write the fractional part of $\pi$ as \[ \pi-\lfloor\pi\rfloor=\sum_{j=1}^{\infty}a_jb^{-j}, \qquad a_j\in\{0,\ldots,b-1\}. \] For a word $w=(w_1,\ldots,w_k)$ of length $k$, count its overlapping occurrences starting at positions $1$ through $N$. Normality in base $b$ means \[ \lim_{N\to\infty}\frac1N \#\{1\le j\le N:(a_j,\ldots,a_{j+k-1})=w\}=b^{-k} \] for every $k$ and every word $w$. The selected conjecture asks for this in every integer base, often called absolute normality. Equal frequencies for individual digits alone are insufficient, as are finite computations of very many digits.

Formulation and concept references: [S1].

Short English statement

In every integer base, do all finite digit patterns appear in pi with exactly the frequencies expected from uniform random digits?

Research attempt

Scope and source check (27 September 2026).

The problem asks about the one specified number $\pi$, in every integer base. Neither the almost-everywhere theorem for real numbers nor computations of a finite prefix identify this number as normal. The formulation source [S1] explains this distinction. Bailey and Crandall [E1] develop equidistribution criteria and conditional routes involving digit-generation recurrences; their dynamical hypothesis is not silently assumed below. This attempt proves a precise sufficient cancellation criterion, the almost-everywhere theorem with overlapping blocks, and explicit obstructions to several tempting shortcuts. The finite computation uses certified rational bounds for $\pi$, not an unverified string of digits.

Lemma 471.1: the exact dynamical reformulation.

Fix $b\geq2$, and use the base-$b$ expansion not ending in an infinite string of digits $b-1$. Let $T_b(x)=\{bx\}$ on $[0,1)$, with braces denoting fractional part. The word $w=(w_1,\ldots,w_k)$ corresponds to \[ a(w)=\sum_{r=1}^k w_r b^{k-r},\qquad J_w=[a(w)b^{-k},(a(w)+1)b^{-k}). \] The block starting at digit $j$ of $\alpha$ is $w$ exactly when $T_b^{j-1}(\{\alpha\})\in J_w$. Consequently base-$b$ normality is equivalent to uniform distribution of \begin{equation}\label{wsref-0-61-34-37-31-3a-6f-72-62-69-74} \{b^j\alpha\},\qquad j=0,1,2,\ldots. \end{equation} For completeness, the implication from normality to uniform distribution is not an assertion that only the listed intervals matter. Given any interval $J\subset[0,1)$, approximate it from inside and outside by unions of intervals of the grid of mesh $b^{-k}$. The difference of their lengths is at most $2b^{-k}$. Normality supplies the limiting frequencies for each finite union. First let the orbit length tend to infinity, and then let $k$ tend to infinity. This sandwiches the frequency in $J$ at its length. The converse follows immediately by taking $J=J_w$.

The choice of expansion causes no problem for $\pi$, which is irrational. More generally the countable set of terminating expansions can be removed when making measure-theoretic assertions.

A concrete analytic target.

Write $e(t)=\exp(2\pi i t)$. The Weyl criterion turns (a471:orbit) into the collection of estimates \begin{equation}\label{wsref-0-61-34-37-31-3a-77-65-79-6c} \frac1N\sum_{j=0}^{N-1}e(hb^j\pi)\longrightarrow0 \quad\text{for every }h\in\mathbb Z\setminus\{0\}. \end{equation} Doing this for every integer $b\geq2$ would solve the selected problem. The criterion follows first for trigonometric polynomials by linearity, then for continuous periodic functions by uniform trigonometric approximation, and finally for interval indicators by upper and lower continuous approximations whose integrals differ arbitrarily little. Conversely uniform distribution gives the integral of each exponential, which is zero for $h\ne0$. Thus (a471:weyl) is an equivalent target, not an extra conjecture claimed to be easier.

The trivial bound for the sum is $N$. A bound that is merely $O(N)$, or a computation for a large but finite $N$, supplies no cancellation in this sense. Nor does irrationality of $\pi$ invoke the elementary equidistribution theorem for $\{j\pi\}$: the exponents here are $b^j$, not $j$. The spacing of this orbit is exactly the difficult information encoded by the digits.

Proposition 471.2: almost every number meets all the required tests.

Under Lebesgue measure on $[0,1)$, the base-$b$ digits are independent and uniform. Indeed prescribing the first $m$ digits specifies a half-open interval of length $b^{-m}$, and summing these cylinder measures establishes independence for arbitrary finitely many digit positions.

Fix a word of length $k$. Let $X_j$ indicate that it begins at position $j$, and $S_N=\sum_{j=1}^N X_j$. Then $\mathbb E X_j=b^{-k}$. Blocks can overlap, but $X_i$ and $X_j$ are independent whenever $|i-j|\geq k$. Bounding each remaining covariance in absolute value by one gives \[ \operatorname{Var}(S_N)\leq(2k-1)N. \] For every $\eta>0$, Chebyshev's inequality implies \[ \sum_{r=1}^{\infty} \mathbb P\left(\left|\frac{S_{r^2}}{r^2}-b^{-k}\right|>\eta\right) \leq\frac{2k-1}{\eta^2}\sum_{r=1}^{\infty}\frac1{r^2}<\infty. \] The first Borel--Cantelli lemma, followed by a countable intersection over positive rational $\eta$, proves convergence along squares almost surely. For $r^2\leq N<(r+1)^2$, monotonicity of $S_N$ or the estimate $|S_N-S_{r^2}|\leq2r+1$ interpolates to all $N$. Finally take the countable intersection over all words, all lengths, and all integer bases. The resulting set of absolutely normal numbers has full Lebesgue measure.

This is a complete theorem with the correct overlapping-block convention. It leaves every particular point in a possible measure-zero exceptional set. A singleton has measure zero, so saying that almost every number is absolutely normal says nothing by itself about whether $\pi$ is among the exceptions. No probability distribution on the already fixed constant changes that logical issue.

Finite block tests cannot imply normality.

For any base $b$ and positive integer $m$, there is a cyclic word of length $b^m$ in which every word of length $m$ occurs exactly once cyclically. Here is the construction. Take the directed graph whose vertices are words of length $m-1$ and whose edges are words of length $m$, directed from their prefix to their suffix. Every vertex has indegree and outdegree $b$, and the graph is strongly connected by successively appending the desired destination digits. A finite balanced connected directed graph has an Euler circuit: follow unused edges until closing a walk, then splice in further closed walks at vertices with unused edges. Reading the last digits of its edges gives the desired cyclic word. For $m=1$ the graph has one empty-word vertex and $b$ loops, with the same conclusion.

Repeat this word forever. In each period every $k$-digit word with $k\leq m$ occurs $b^{m-k}$ times, so all those word frequencies are exactly $b^{-k}$. But the period has only $b^m$ starting positions and there are $b^{m+1}$ possible words of length $m+1$. Some such words never occur. The expansion is periodic and is not normal.

Prepending any specified finite prefix does not alter any limiting block frequency. Consequently every finite prefix can be followed by a tail that passes every word-frequency test up to any specified finite length $m$ but fails normality at a longer length. Conversely, every finite prefix cylinder has positive measure, so Proposition 471.2 says it contains absolutely normal numbers. Thus finite digits alone, even perfectly balanced finite-length tests, do not logically decide normality.

Transcendence and simple normality still do not suffice.

There is an explicit stronger diagnostic. Start with the periodic base-$b$ number \[ \alpha_0=0.\overline{012\cdots(b-1)}{}_b. \] Choose $j_0\geq\max\{b,3\}$, and change its digit at each position $j!$, $j\geq j_0$, from $b-1$ to zero. These positions are multiples of $b$. The resulting number is exactly \begin{equation}\label{wsref-0-61-34-37-31-3a-6c-69-6f-75-76-69-6c-6c-65} \alpha=\alpha_0-(b-1)\sum_{j=j_0}^{\infty}b^{-j!}. \end{equation} The operation changes the displayed digits directly and causes no borrowing: each indicated digit of $\alpha_0$ is $b-1$. The number of changed positions up to $N$ is $o(N)$. Changing one digit alters at most $k$ occurrences of any fixed $k$-digit word. Hence all limiting word frequencies of $\alpha$ equal those of the periodic background. In particular every single digit has frequency $1/b$, but the word $00$ has frequency zero instead of $1/b^2$. The number is simply normal and fails normality in base $b$.

It is nevertheless transcendental. It is first irrational: if its digits were eventually periodic, comparison with the periodic background would make the set of disagreements eventually periodic. An infinite eventually periodic subset of the integers has positive density, contradicting the infinite, density-zero set of factorial positions. The base expansion is unambiguous here.

Truncating the series in (a471:liouville) at $j=s$ gives a rational $p_s/q_s$ with \[ q_s\leq(b^b-1)b^{s!},\qquad 0<\left|\alpha-\frac{p_s}{q_s}\right| \leq2(b-1)b^{-(s+1)!}. \] For every fixed $A>0$, this is less than $q_s^{-A}$ for sufficiently large $s$, since $(s+1)!=(s+1)s!$. To see why that excludes algebraicity, suppose a nonzero integer polynomial $P$ of degree $d$ vanishes at $\alpha$. For sufficiently close rational approximants distinct from its finitely many roots, \[ |P(p/q)|\geq q^{-d},\qquad |P(p/q)-P(\alpha)|\leq C|p/q-\alpha| \] by integrality after multiplication by $q^d$ and a derivative bound on a fixed compact interval. Thus $|p/q-\alpha|\geq C^{-1}q^{-d}$, contradicting the factorial approximants. This proves the claimed transcendence without assuming any digit-randomness principle.

The example does not model the special analytic structure of $\pi$. Its role is narrower and rigorous: even transcendence together with correct one-digit frequencies leaves the desired higher-block conclusion unproved.

Certified finite experiment.

For the finite check, Machin's identity gives \[ \pi=16\arctan(1/5)-4\arctan(1/239). \] It can be verified by the tangent addition formula: $\tan(4\arctan(1/5))=120/119$, and subtracting $\arctan(1/239)$ gives tangent one, with the angle in the correct first-quadrant branch. Alternating rational bounds for \[ \arctan z=\sum_{r=0}^{\infty} \frac{(-1)^r z^{2r+1}}{2r+1} \] therefore bound $\pi$ between explicit rational endpoints. Taking 256 terms for each series gives enough precision for the following small diagnostic. For each $b\in\{2,3,10,16\}$ the script checks that the two endpoints give the same integer part after multiplication by $b^{260}$. This equality certifies the first 260 fractional digits, after subtracting the integer part three and converting the remainder to base $b$.

For every word length $1\leq k\leq4$, the diagnostic counts the blocks beginning at positions $1,\ldots,256$, using the certified extra digits for the last overlapping blocks. It records the complete count table and checks that the counts sum to 256. These are finite identities, not limits. The bounds establish that the digits counted really belong to $\pi$; they do not turn a finite frequency deviation into an asymptotic discrepancy estimate. The same diagnostic constructs the cyclic words above and checks their exact short-word frequencies and missing longer words.

The remaining analytic gap.

An effective digit algorithm provides digits at chosen places or a finite precision approximation. To prove normality it must additionally yield cancellation such as (a471:weyl), uniformly as the length tends to infinity for each fixed base and frequency. The recurrence-based hypotheses discussed in [E1] propose such additional information; treating a hypothesis as an available theorem would simply move the unanswered question into an assumption. None of the arguments above bounds those sums by $o(N)$ for $\pi$ in even one fixed base.

The proved statements locate the boundary of the attempt: the exact dynamical equivalence, the full-measure theorem, explicit transcendental nonnormal examples, and certified finite digits are available. A theorem selecting $\pi$ from the full-measure normal set, or proving the equivalent cancellation for its lacunary orbit in all bases, is absent. Absolute normality of $\pi$ remains unresolved here.

UNRESOLVED. The partial results above do not establish the full selected statement.

Sources

• S1 Davar Khoshnevisan, "Normal Numbers are Normal," Clay Mathematics Institute Annual Report 2006, pp. 15, 27-31. https://www.claymath.org/library/annual_report/ar2006/06report_normalnumbers.pdf. Formulation source.
• E1 D. H. Bailey and R. E. Crandall, "Random Generators and Normal Numbers," author-hosted primary paper. https://www.davidhbailey.com/dhbpapers/bcnormal.pdf. Equidistribution criteria and explicitly conditional dynamical routes; checked 27 September 2026.