Mahler's 3/2 problem · a computation, on one laptop

No Mahler Z‑number below 2⁶³

In 1968 Kurt Mahler asked whether some positive number ξ keeps the fractional part of ξ, ξ·3/2, ξ·(3/2)², ξ·(3/2)³, … below one half, for ever. Such a number is called a Z-number. Nobody knows whether one exists, and Mahler guessed that none does. We checked every positive number below 2⁶³, about 9.2 quintillion. None of them passes. In 2009 Dubickas and Mossinghoff showed by computer that a Z-number, if there is one, must exceed 2⁵⁷. The census on this page moves that floor 64 times higher, and it took a quarter of an hour on a laptop.

Seen

The longest survivor below 2⁶³, the integer part n = 1541211973603965806. The top strip is its parity, black where the integer part of ξ·(3/2)^j is odd. The dots are the fractional parts of a candidate ξ* chosen exactly from the last window of survivors (the middle of it): every one sits below the dashed line at ½ for j = 0 to 156, and at j = 157 it jumps above, as does every other candidate in [n, n+1). Computed exactly in your browser.

The record horizon among all starts below 2^w, for w from 12 to 63 bits (dots), against the coin law of about 2.41·w steps (line). The records follow chance closely: nothing below 2⁶³ lasts anomalously long.

Stated

The result (computed, exact)

For every integer n with 1 ≤ n < 2⁶³, no real ξ in [n, n+1) has the fractional part of ξ·(3/2)^j below ½ for every j ≥ 0. Call the horizon of n the first j at which no ξ in [n, n+1) is left below one half: every start below 2⁶³ has a finite horizon, at most 157. A number in (0, 1) fails too: its multiples ξ·(3/2)^j grow by half each step, so one of them lands between ½ and 1. So there is no Z-number in (0, 2⁶³).

Why there is one candidate per integer (Mahler 1968; proved)

Write ξ·(3/2)^j = n_j + u_j, with n_j an integer and u_j in [0, ½). Then the integer parts follow one fixed map, n_j+1 = ⌈3n_j/2⌉. The fractional part is fixed by the future parities b_k = n_k mod 2 alone:

u_j = Σ_k≥0 b_j+k · 2^k / 3^k+1

So the integer part n₀ decides everything. Each unit interval [n, n+1) holds at most one Z-number, and the question becomes whether the orbit of some positive integer under ⌈3x/2⌉ keeps every one of those sums below ½.

How 9.2 quintillion candidates are checked in minutes

The first T parities of an orbit depend only on n mod 2^T, its lowest T binary digits. So every start that shares those digits shares its early life. The census walks the tree of low digits. Along each branch it keeps the exact interval of fractional parts still alive, as integers over a power of two. When an interval empties, every start below that branch dies at once, sometimes trillions together. A start that survives all 63 digits runs on alone until it fails. The 4,499 starts still alive after 124 steps were replayed in exact rational arithmetic; the longest horizon was 157. The total is checked: exactly 2⁶³ − 1 starts, each with a finite horizon.

How it was checked

Before the full run, the census matched an independent exact-fraction computation on every start below 2²⁰. The full run was repeated, with the same 4,499 stragglers and the same record. Our reviewing model, GPT, audited the source and recomputed fixed cases independently. Dubickas and Mossinghoff's 2009 table lists one of our record starts, 9850069454682037, which fails at step 137, so the two searches use the same definitions. One more cross-check came out of the counts: the number of surviving digit patterns of each length equals, exactly, the word count of the base-3/2 digit system (the greedy beta-shift), at every length below 63.

Starts below	Record horizon	Its integer part
2²⁰	47	690773
2³²	76	3097089134
2⁴⁰	99	704731028141
2⁴⁸	120	119619616250377
2⁵⁶	137	9850069454682037
2⁶⁰	143	439797727552450048
2⁶³	157	1541211973603965806

Sources. MAHLER.md section 1 (the census and its receipts); mahler_census.c and mahler_census.py (the walker, its predictions, controls and outcome); RULE30-GPT.md GC1221 (GPT's audit) and CHAT-LEDGER.15.md GC1225 (the published-table cross-check); A. Dubickas and M. J. Mossinghoff, Lower bounds for Z-numbers, Math. Comp. 78 (2009) 1837–1851; K. Mahler, An unsolved problem on the powers of 3/2, J. Austral. Math. Soc. 8 (1968) 313–321.

The other half

Mahler's question is usually asked for positive ξ. For a negative ξ the same question is equivalent to asking for a positive η whose fractional parts all stay in the upper half (½, 1), because the map n ↦ −1 − n turns one problem into the other: it commutes with ⌈3x/2⌉ and swaps every parity. Akiyama, Frougny and Sakarovitch state the question for every nonzero ξ, and we found no published treatment of this half on its own.

The negative starts (computed, exact)

The same census run on every negative integer part x with −2⁶³ < x ≤ −1: every one has a finite horizon, the longest 153, at x = −6964446546082240544. So no positive η up to 2⁶³ − 1 keeps every fractional part of η·(3/2)ⁿ in (½, 1). As the symmetry requires, the counts over the first 63 steps match the positive side exactly. The records and the tail follow the same chance law: 4,622 starts alive at step 124, against 4,499 on the positive side.

Checked, here, now

Type any positive integer n, or press one of the record starts. Your browser follows its orbit under ⌈3x/2⌉ with exact integer arithmetic and reports the step at which no ξ in [n, n+1) is left below one half.

The start 4 is the small case worth watching: some ξ in [4, 4½) keeps below one half for j = 0 to 11, a horizon of 12. Starts 1, 2, 3 and 5 have horizons 4, 3, 2 and 6.

Compute is the only limit

This ran on a single Apple M5 MacBook Pro, in about a quarter of an hour (798 seconds on 9 cores). The work grows by about half again for each extra binary digit: 2⁶⁰ took 237 seconds and 2⁶³ took 798, a ratio of 3.37 against (3/2)³ = 3.375. Extrapolating:

Bound	One laptop (estimate)	A thousand cores (estimate)
2⁶³ (done)	13 minutes	seconds
2⁷⁰	about 4 hours	about 2 minutes
2⁸⁰	about 9 days	about 2 hours
2⁹⁰	about 1.5 years	about 5 days

The walk splits by the lowest digits into independent pieces, so it spreads over any number of machines with no communication. Going beyond 2⁶³ needs wider integers: start values past 64 bits, and past about 2⁸⁰ wider intermediate values in the walker too. If you have the machines, take the code, run it, and send us the receipt: the histogram, the record and the stragglers. We will replay it. Details on the join page.

What it is and what it is not

It is an exact finite verification, and it extends a published bound by a factor of 64. It is also good evidence of a mild kind: the record survivors grow like chance, about 2.4 steps per binary digit, and nothing below 2⁶³ behaves strangely. It is not a proof. Any finite range can be checked, but "for ever" cannot be reached this way: Turing's 1936 point, that finitely many checks never settle a question about all numbers without an essentially new argument. A Z-number could still exist far above any bound a computer will ever reach. We are not calling 2⁶³ the current record either: a brief search found no improvement on 2⁵⁷ since 2009, but we have not combed the later literature. The census is one party's computation, audited at the source by a second, not yet replayed in full by anyone else.

Sources. MAHLER.md section 1 and PRIOR-ART.md ("Mahler's 3/2 problem: survey at the room switch", with its correction crediting Dubickas and Mossinghoff).