Erdős problem #269

Three-prime running least common multiples

For a finite set of at least two primes, is the sum of reciprocals of the running least common multiples of the smooth numbers irrational? This library treats the three-prime case.

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The problem

The question

Choose a finite set of at least two primes. List 1 and every positive integer made from those primes, in increasing order. After each number, find the least common multiple (LCM): the smallest positive integer divisible by everything listed so far. Add its reciprocal, even when this running value repeats. Must the infinite sum be irrational, meaning it cannot be written as a fraction of two integers?

An example

Allow 2, 3 and 5. The list starts $1, 2, 3, 4, 5, 6, \dots$ ; the running multiples are $1, 2, 6, 12, 60, 60, \dots$ . Adding 5 changes 12 to 60. Adding 6 leaves 60 unchanged, since 6 already divides 60. Both numbers contribute: the first six terms give $1 + 1 / 2 + 1 / 6 + 1 / 12 + 2 / 60 = 107 / 60$ .

Let $P$ be a finite set of primes with $| P | \geq 2$ and let ${a_{1} < a_{2} < \dots} = {n \in N : if p ∣ n then p \in P}$ . Is the sum $\sum_{n = 1}^{\infty} \frac{1}{[a_{1}, \dots, a_{n}]},$ where $[a_{1}, \dots, a_{n}]$ is the lowest common multiple of $a_{1}, \dots, a_{n}$ , irrational?

Erdős problem #269, as erdosproblems.com/269 states it

The setup in detail

Read the statement symbol by symbol. $P$ is a finite set of primes, at least two of them, such as ${2, 3}$ or ${2, 3, 5}$ . The set in braces holds every positive whole number $n$ whose prime divisors all belong to $P$ . It is infinite, since it contains every power of each prime in $P$ , and $a_{1} < a_{2} < a_{3} < \dots$ lists it in increasing order.

The bracket is the least common multiple: the smallest positive whole number that each of $a_{1}, \dots, a_{n}$ divides. For example $[4, 6] = 12$ . It is a running least common multiple, recomputed each time one more number joins, so the sum has one term for every $n$ , and equal terms are all counted.

Irrational means it cannot be written as $A / D$ with integers $A, D$ and $D > 0$ . A proof must rule out every denominator $D$ at once, and no amount of computed decimal places can do that.

The question is asked for every finite set of at least two primes. With one prime the sum is rational: for $P = {2}$ the list is $1, 2, 4, 8, \dots$ , the running least common multiple of the first $n$ numbers is $2^{n - 1}$ , and

1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = 2.

For an infinite set of primes, Erdős (1988) called irrationality a simple exercise. Two primes are settled, as the crux explains; three or more primes are open.

The example in detail

The next numbers are $8, 9, 10, 12$ . Including $8 = 2^{3}$ requires a third factor 2, changing 60 to $60 \times 2 = 120$ . Including $9 = 3^{2}$ changes 120 to 360; 10 and 12 already divide 360. Thus the first ten terms, grouped over the common denominator 360, are:

1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{2}{60} + \frac{1}{120} + \frac{3}{360} = \frac{360 + 180 + 60 + 30 + 12 + 3 + 3}{360}

= \frac{648}{360} = \frac{9}{5} .

Counting each distinct multiple once instead gives

1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{60} + \frac{1}{120} + \frac{1}{360} = \frac{640}{360} = \frac{16}{9} .

The two differ by $8 / 360 = 1 / 45$ : the second copy of $1 / 60$ and the extra two copies of $1 / 360$ .

The diagram's last stretch, from 16 to just below 32, holds 16, 18, 20, 24, 25, 27, 30, with running least common multiples 720, 720, 720, 720, 3600, 10800, 10800, so this stretch adds $4 / 720 + 1 / 3600 + 2 / 10800$ . All sixteen numbers of the list from 256 to 500 share the running least common multiple $2^{8} \times 3^{5} \times 5^{3} = 7776000$ .

The sum is finite. From $2^{a}$ to just below $2^{a + 1}$ the list holds at most $(a + 1)^{2}$ numbers: the exponents of 3 and 5 are each at most $a$ , giving $(a + 1)^{2}$ choices, and each choice leaves room for at most one power of 2, since doubling leaves the range. Each of their terms is at most $1 / 2^{a}$ , because $2^{a}$ is already in the list. These bounds add up to $\sum_{a \geq 0} (a + 1)^{2} / 2^{a} = 12$ . Computed exactly below $2^{40}$ , with the same bound for the rest, the sum is 1.8093 to four decimal places.

The crux

The diagram continues this example to 30. The staircase shows the running multiple; each rise multiplies it by the labelled prime. Red flat stretches mark repeated values. The bottom row groups their contributions: four numbers sharing 720 add $4 / 720$ . Counting each value once loses these repeats and changes the question. Steve Fan settled the two-prime question. For three primes, the result below rules out a finite factorisation extending his method; the original irrationality question stays open.

Why the staircase rises only at prime powers. Up to 24 the largest allowed prime powers are $16, 9, 5$ , whose product is $16 \times 9 \times 5 = 720$ . Every allowed number up to 24 divides this product: for example, $24 = 8 \times 3$ . Conversely, the running LCM must contain each of those three powers. Generally it is the product of the largest available power of each prime. A new prime power adds one factor of its prime; other numbers leave it unchanged. The staircase's horizontal spacing counts listed numbers, rather than measuring their numerical distance.

A different sum that is settled. Suppose we count each distinct running value once. After the value L, multiply the remaining sum, its tail, by L. Each subsequent value is at least twice the preceding one, so this scaled tail lies between 0 and 1. If the complete sum were A/D, D times each scaled tail would be an integer: all earlier denominators divide L. Only finitely many scaled tails could then occur.

For primes 2, 3 and 5, the short paper proves that a scaled tail determines both the next block of prime-power jumps and the next tail. Finitely many states would therefore force eventual periodicity, just as finitely many remainders force repeating digits in long division. If each repeating block spanned $ℓ$ doublings and contained $c$ powers of 3, their long-run frequencies would require $2^{ℓ} = 3^{c}$ , impossible. This proves irrationality for the distinct-value sum. Erdős asserted the conclusion for every finite set of at least two primes in a 1973 letter without printing a proof; the paper claims no priority for this conclusion.

Why repeats matter. For the original sum, multiply the contribution from 16 to just below 32 by the preceding running value 360:

4 \cdot \frac{360}{720} + \frac{360}{3600} + 2 \cdot \frac{360}{10800} = \frac{13}{6} .

The scaled tails are no longer below 1. The long paper proves that those taken at powers of 2 grow quadratically with the exponent. Under rationality, D times these tails would still be integers, but unbounded integers need not repeat. The distinct-value proof therefore does not settle the original sum.

The factorisation route. Steve Fan proved transcendence, and hence irrationality, for every pair of distinct primes by factoring the repeated sum into two series and applying a theorem of Loxton and van der Poorten. The result below rules out extending the table factorisation to three distinct primes by any finite sum of separated products. It does not rule out a different proof. The original question remains open for finite sets of three or more primes.

Make a table: rows 1, 2, 4, …; columns 1, 3, 9, …. Each cell is the reciprocal running LCM at its labels' product. Allowing only primes 2 and 3 makes row 2 half of row 1. Allow prime 5 too. At cutoff 6, the LCM of 1, 2, 3, 4, 5, 6 is 60. Copying the first row predicts $1 / 2 \times 1 / 6 = 1 / 12$ ; the actual cell is $1 / 60$ . Divide cells by their predictions from the first row and column. The corner becomes 1, 1 above 1, $1 / 5$ . These are the only values anywhere: blue means the prediction holds; red means an extra factor 5. The strips share a 1-to-5 axis. A row marker removes full powers of 5: row 8 gives 8/5. A column probe marks the factor reaching its next power of 5: column 3 gives 5/3. Reaching a probe makes that cell red. Subtract neighbouring strips. Their red bar is nonzero between their markers. One probe in each gap tests that difference alone; the probe at 5 tests the retained first row. These tests establish independent rows. Insert row 8: its marker $8 / 5$ splits the gap from 1 to 2. Column 81's probe $125 / 81$ tests the first piece; column 3's probe $5 / 3$ tests the second. The proof supplies columns in every gap. Thus independent rows never run out. For three distinct primes, the table cannot be written as a finite sum of a row factor times a column factor. Irrationality of the sum remains open.

Let H(x) be the LCM of the positive integers at most x whose prime factors lie in {2,3,5}, and put $K (i, j, k) = 1 / H (2^{i} 3^{j} 5^{k})$ for nonnegative exponents. Fixing k gives a matrix. A separated representation means a finite sum of terms $f_{ℓ} (i) G_{ℓ} (j, k)$ : each term separates the row exponent from the other two. With d terms, every row is a combination of d fixed row patterns. Thus the matrix has rank at most d, and a nonzero determinant of order d+1 excludes it.

The prediction is $P_{i j} = K (i, 0, k) K (0, j, k) / K (0, 0, k)$ . Division by P multiplies each row and column by a nonzero scalar and preserves rank. Writing $A_{i} = 2^{i} / 5^{⌊ \log_{5} 2^{i} ⌋}$ and $S_{j} = 5^{⌊ \log_{5} 3^{j} ⌋ + 1} / 3^{j}$ , the normalised matrix is $C_{i j} = 1 - \frac{4}{5} 1_{{S_{j} \leq A_{i}}}$ , independent of k. The indicator is 1 when its condition holds and 0 otherwise. Its condition means that the remaining row and column factors together cross a power of 5, adding one factor 5 to H and dividing the reciprocal by 5.

For $A_{u} < A_{v}$ , subtraction gives $C_{v j} - C_{u j} = - \frac{4}{5} 1_{{A_{u} < S_{j} \leq A_{v}}}$ . The differences are nonzero on disjoint intervals. Select a column in each and the reference column j=0, whose threshold is 5. Evaluating a linear relation on these columns first kills every difference coefficient and then the retained-row coefficient. This proves independence.

The drawing has four rows and determinant $64 / 125$ ; that finite calculation illustrates the construction. The proof for every order uses the irrationality of $\log_{5} 3$ and density of ${j \log_{5} 3}$ , where braces mean fractional part: thresholds therefore enter every open interval in $(1, 5)$ . Distinct row markers allow arbitrarily many disjoint intervals. Theorem B.2 proves the general statement for any three distinct primes, with indices independent of k. It excludes finite exact separation. The original, decaying kernel can still be approximated by finite separated sums; neither statement settles irrationality.

Results and open questions

#269. For three pairwise distinct primes, the threshold-column argument produces nonsingular selected kernel minors of every order and excludes every finite rational separated kernel representation. The arbitrary-order theorem is present in the compiled public Lean source, but the Wave-A receipt’s named axiom audit does not audit that rank declaration; Comparator checks only the displayed rank-two minor $- 1 / 15$ . For the actual ${2, 3, 5}$ series, Lean checks the rationality-to-positive-reduced-carry bridge. Cofinal local-window escape then extinguishes such carries; under the explicit cap hypotheses recorded in the claim registry, that escape statement is equivalent to irrationality. This is an exact interface, not an easier theorem whose premise has been proved. The source-specific escape and three-prime irrationality remain open.

For every pair of distinct primes, both the repeated running-LCM reciprocal sum and the version retaining one term at each distinct LCM value are transcendental. The ordinary proof expresses them as nonconstant quadratic and affine polynomials in one transcendental Hecke–Mahler value. Steve Fan posted the repeated-sum two-prime factorisation, reduction and transcendence conclusion first, on the problem’s forum on 26 June 2026; the note credits his priority, gives a proof found independently, derives the de-duplicated formula, and uses the cited Hecke–Mahler transcendence theorem. Lean also checks the two-prime affine and quadratic formulas and derives transcendence from the cited Hecke–Mahler result as an explicit named hypothesis; that external theorem itself is not formalised here. Read the short paper, long record, and selected formal checks.

Read the public result guide

Short paper, 16 pages

Distinct running least common multiples

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In this paper

Two sums from running least common multiples

Let $P$ be a finite set of primes and $S_{P}$ the positive integers with prime factors in $P$ , including $1$ . For $P = {2, 3, 5}$ , the running least common multiples of these $5$ -smooth integers begin $1, 2, 6, 12, 60, 60, 120, 360, 360, \dots$ . Put

L (x) = lcm {u \in S_{P} : u \leq x}, H_{P} (x) = \prod_{p \in P} p^{⌊ \log_{p} x ⌋} (x \geq 1) .

Every member of the prefix divides $H_{P} (x)$ , while the prefix contains the largest power of each $p \in P$ at most $x$ . Thus $L (x) = H_{P} (x)$ , with a jump by $p$ exactly at each positive power of $p$ . Counting distinct running LCMs or retaining multiplicities gives

D_{P} = 1 + \sum_{\begin{matrix} t = p^{n} \\ p \in P, n \geq 1 \end{matrix}} \frac{1}{H_{P} (t)}, R_{P} = \sum_{u \in S_{P}} \frac{1}{L (u)} .

For $P = {2, 3, 5}$ , $L (5) = L (6) = 60$ contributes $1 / 60$ to $D_{P}$ and $2 / 60$ to $R_{P}$ . Erdős Problem #269 asks about $R_{P}$ [1] and [2]. Positive powers of distinct primes never coincide; geometric growth of successive running LCMs gives convergence of $D_{P}$ . We use $N = {0, 1, 2, \dots}$ .

Theorem 1.1 (the sum over distinct running LCMs for ${2, 3, 5}$ ). The number $D_{{2, 3, 5}}$ is irrational.

Rationality would leave only finitely many normalised tails, but finiteness alone does not force the jumps to repeat. The decisive step is to recover the next block and the following tail from the current tail. Powers of $2$ delimit the blocks; the spacing of powers of $5$ separates their five affine images. Equal tails then have identical futures, contradicting the irrational frequency of powers of $3$ .

Erdős asserted irrationality of $D_{P}$ for every finite $P$ with $| P | \geq 2$ in a letter dated 1 January 1973, without printing a proof [3]. Thus the theorem above is a special case of that earlier assertion. We claim no priority for its conclusion. The companion’s general argument uses an isolated interchange of jumps on a torus. That argument is retained there as an ordinary proof awaiting formalisation. The proof below is independent of that argument.

Our arguments leave the repeated sum $R_{P}$ unresolved for finite $P$ with $| P | \geq 3$ . For $P = {p}$ both sums equal $p / (p - 1)$ .

After the five-map proof in Section 2, we retain the multiplicities. The bases remain the same, but the block coefficients and tails grow quadratically. Sections 3 and 4 replace finite-state repetition by a test for excluding integer tails. For two primes, Fan’s factorisation [7] reduces both sums to a Hecke–Mahler value and yields transcendence (Section 5). The appendices supply the running LCM identities and show why exact finite separation of the three-prime kernel cannot extend that factorisation.

Recovering the prime-power jumps

Long division combines a finite set of remainders with a deterministic rule for the next digit and remainder. Here the bases vary with the prime-power jumps, so that rule must be proved. We first obtain finitely many tails under rationality, then recover the jumps from each tail.

Proof of Theorem 1.1. Finitely many tails under rationality. Write $H = H_{{2, 3, 5}}$ in this proof and put $P_{a} = H (2^{a})$ . Every running LCM at or before $2^{a}$ divides $P_{a}$ . We therefore normalise the tail after $2^{a}$ by

\begin{matrix} (1) & Y_{a} = P_{a} \sum_{\begin{matrix} t > 2^{a} \\ t = 2^{n}, 3^{n} or 5^{n}, n \geq 1 \end{matrix}} \frac{1}{H (t)} . \end{matrix}

Each subsequent jump multiplies the running LCM by at least $2$ , and a jump by $3$ eventually occurs. Comparison with $\sum_{j \geq 1} 2^{- j}$ gives $0 < Y_{a} < 1$ .

Suppose that $D_{{2, 3, 5}} = N / K$ , where $N$ is an integer and $K \geq 1$ is an integer. Clearing the finite prefix gives

K Y_{a} = P_{a} N - K (P_{a} + \sum_{\begin{matrix} t \leq 2^{a} \\ t = 2^{n}, 3^{n} or 5^{n}, n \geq 1 \end{matrix}} \frac{P_{a}}{H (t)}) \in Z .

Thus the $Y_{a}$ lie in the finite set $(K^{- 1} Z) \cap (0, 1)$ . It remains to show why equality of two tails propagates.

Recovering a block from its tail. Consider the jumps between $16$ and $32$ , including the last endpoint. They occur at $25, 27, 32$ , multiplying the running LCM successively by $5, 3, 2$ . After normalisation at $16$ , their contribution is $1 / 5 + 1 / 15 + 1 / 30$ . The remaining tail is normalised at a running LCM $30$ times as large, so

Y_{4} = \frac{1}{5} + \frac{1}{15} + \frac{1}{30} + \frac{Y_{5}}{30} = \frac{9 + Y_{5}}{30} .

In the block $(64, 128]$ , the interior jumps occur in the opposite order, at $81, 125$ . Its constant term is $1 / 3 + 1 / 15 + 1 / 30 = 13 / 30$ . The multiplier $30$ records which primes occur, but not their order. The numerators $9$ and $13$ retain that order. To read it from a tail, we must also control the unknown following tail.

Every block $(2^{a}, 2^{a + 1}]$ contains its terminal power of $2$ , at most one power of $3$ and at most one power of $5$ . We record the interior jumps in their order by a word $τ_{a}$ on the letters $3, 5$ , omitting the terminal $2$ . Splitting off the block gives

\begin{matrix} (2) & Y_{a} = G_{τ_{a}} (Y_{a + 1}), G_{τ} (y) = \frac{μ_{τ} + y}{b_{τ}}, \begin{array}{cccccc} τ & \emptyset & 3 & 5 & 35 & 53 \\ b_{τ} & 2 & 6 & 10 & 30 & 30 \\ μ_{τ} & 1 & 3 & 3 & 13 & 9 \end{array} \end{matrix}

Here $b_{τ}$ is the factor by which the running LCM increases across the block, and $μ_{τ} / b_{τ}$ is the block’s normalised contribution.

The bound $0 < Y_{a + 1} < 1$ does not yet recover $τ_{a}$ : $G_{53} ([0, 1]) = [3 / 10, 1 / 3]$ and $G_{5} ([0, 1]) = [3 / 10, 2 / 5]$ overlap. The table improves the lower bound to $Y_{a} \geq 3 / 10$ . To separate these two images, an upper bound $U$ must satisfy

G_{53} (U) < G_{5} (3 / 10), \frac{9 + U}{30} < \frac{33}{100}, U < \frac{9}{10} .

The spacing of powers of $5$ supplies such a bound. A power $5^{f}$ belongs to block $⌊ f \log_{2} 5 ⌋$ . These indices start at $2$ and their successive differences are $2$ or $3$ . Every three consecutive blocks therefore contain a power of $5$ . Given $a$ , let $j \in {a, a + 1, a + 2}$ be the first such block. The maps for blocks containing $5$ give $Y_{j} \leq 7 / 15$ , since $Y_{j + 1} < 1$ and $G_{35} (1) = 7 / 15$ bounds $G_{5} (1)$ and $G_{53} (1)$ as well. We now work backwards from $j$ to $a$ . Each intervening block has map $G_{\emptyset}$ or $G_{3}$ , and $G_{3} (y) \leq G_{\emptyset} (y) = (1 + y) / 2$ for $y \geq 0$ . At most two backward steps give the successive bounds $7 / 15$ , $11 / 15$ , $13 / 15$ . Hence

\begin{matrix} (3) & Y_{a} \in I = [\frac{3}{10}, \frac{13}{15}], G_{\emptyset} (G_{\emptyset} (7 / 15)) = \frac{13}{15} < \frac{9}{10} . \end{matrix}

This bound holds for the tails of the prime-power sequence because every three consecutive blocks contain a power of $5$ . The interval $I$ is not invariant under arbitrary compositions of the five maps: for example, $G_{\emptyset} (13 / 15) = 14 / 15$ lies outside $I$ .

Figure 1 gives the exact images of $I$ . They are pairwise disjoint, with successive gaps $1 / 900, 17 / 300, 79 / 900, 1 / 180$ . Writing $L = 3 / 10$ and $U = 13 / 15$ , the two narrow gaps are

G_{5} (L) - G_{53} (U) = \frac{3 L - U}{30} = \frac{1}{900}, G_{\emptyset} (L) - G_{3} (U) = \frac{3 L - U}{6} = \frac{1}{180} .

Thus the same inequality $U < 3 L = 9 / 10$ separates both pairs.

This figure is drawn in the PDF.

Each actual $Y_{a}$ lies in exactly one image interval, identifying $τ_{a}$ and then $Y_{a + 1} = b_{τ_{a}} Y_{a} - μ_{τ_{a}}$ . For instance, $Y_{a} \in [31 / 100, 74 / 225]$ forces $τ_{a} = 53$ and $Y_{a + 1} = 30 Y_{a} - 9$ . This is the next arithmetic tail, not an arbitrary inverse image; it lies in $I$ , so recovery can be repeated.

Repetition would give a rational frequency. The finitely many tail values obtained above include a repeated value $Y_{u} = Y_{v}$ with $u < v$ . Since the image intervals are disjoint, this forces $τ_{u} = τ_{v}$ . Applying the corresponding inverse map gives $Y_{u + 1} = Y_{v + 1}$ . Induction yields $τ_{u + j} = τ_{v + j}$ for every $j \geq 0$ . Thus the entire later block word has period $v - u$ .

To exclude this possibility, let $α = \log 2 / \log 3$ . The number of powers of $3$ in block $a$ is $⌊ (a + 1) α ⌋ - ⌊ a α ⌋$ , and is determined by $τ_{a}$ . If the block word has eventual period $ℓ \geq 1$ , let $c$ be the number of $3$ -powers in one period. For every sufficiently large fixed $a$ and every $n \geq 1$ we then have

⌊ (a + n ℓ) α ⌋ = ⌊ a α ⌋ + n c .

Dividing by $n$ and taking the limit gives $ℓ α = c$ , contrary to unique factorisation, since it would imply $2^{ℓ} = 3^{c}$ . ◻

The repeated three-prime sum

We now retain every smooth integer, not just the jumps of its running LCM. Put $P = {2, 3, 5}$ , $S = R_{P}$ and $H = H_{P}$ throughout this section. The denominator changes at the same prime powers; the number of terms at each denominator is what changes the argument.

For example, $[16, 32)$ contains the seven smooth integers $16, 18, 20, 24, 25, 27, 30$ . Grouping them by their running LCM gives

\begin{array}{ccc} smooth integers & running LCM & contribution \\ 16, 18, 20, 24 & 720 & 4 / 720 \\ 25 & 3600 & 1 / 3600 \\ 27, 30 & 10800 & 2 / 10800 \end{array}

The distinct sum counts each displayed running LCM once; the repeated sum assigns the coefficients $4, 1, 2$ . The jumps alone therefore no longer determine the block contribution. Appendix A gives the general finite grouping identity.

With blocks $[2^{a}, 2^{a + 1})$ , the tail now includes $2^{a}$ . The prefix excludes the jump by $2$ there, so half the running LCM clears it. Put

s_{a} = \sum_{\begin{matrix} x \in S_{P} \\ 2^{a} \leq x < 2^{a + 1} \end{matrix}} \frac{1}{H (x)}, T_{a} = \sum_{j \geq 0} s_{a + j}, h_{a} = \frac{H (2^{a})}{2}, X_{a} = h_{a} T_{a} .

For $a \geq 1$ , each prefix LCM has exponent of $2$ at most $a - 1$ and divides the integer $h_{a}$ ; at $a = 0$ , $h_{0} = 1 / 2$ . The factor $1 / 2$ reflects the endpoint change from Section 2, not multiplicity.

Splitting $T_{a} = s_{a} + T_{a + 1}$ and using $h_{a + 1} = b_{a} h_{a}$ gives $X_{a + 1} = b_{a} X_{a} - m_{a}$ , where $m_{a} = h_{a + 1} s_{a}$ . Explicitly,

\begin{matrix} (4) & b_{a} = \frac{H (2^{a + 1})}{H (2^{a})}, m_{a} = \sum_{\begin{matrix} x \in S_{P} \\ 2^{a} \leq x < 2^{a + 1} \end{matrix}} \frac{H (2^{a + 1})}{2 H (x)} . \end{matrix}

Since $h_{0} = 1 / 2$ , we also have $h_{a + 1} = (b_{0} \dots b_{a}) / 2$ and $m_{a} / (b_{0} \dots b_{a}) = s_{a} / 2$ . For $[2, 4)$ , the running LCMs at $2, 3$ are $2, 6$ and $h_{2} = 6$ , giving $m_{1} = 6 (1 / 2 + 1 / 6) = 4$ and $b_{1} = 6$ . For the block displayed above, $h_{5} = 10800$ and

m_{4} = 4 \cdot 15 + 3 + 2 \cdot 1 = 65, b_{4} = 30.

Lemma 3.1 (integer coefficients and four possible bases). For every $a \geq 0$ , $m_{a}$ is a positive integer and $b_{a} \in {2, 6, 10, 30}$ . The word “numerator” does not impose the positional-digit restriction $m_{a} < b_{a}$ ; that restriction need not hold.

Proof. For $x < 2^{a + 1}$ we have $2 H (x) ∣ H (2^{a + 1})$ , so every summand in $m_{a}$ is integral. The interval contains $2^{a}$ , giving positivity. Between successive powers of $2$ there is at most one power of $3$ and at most one power of $5$ . Along with the final factor $2$ , these give $b_{a} \in {2, 6, 10, 30}$ . ◻

Proposition 3.2 (the tail recurrence and a quadratic bound). The series defining $S, T_{a}$ converge. For every $a \geq 0$ ,

X_{a + 1} = b_{a} X_{a} - m_{a}, 0 < X_{a} \leq \frac{8640}{343} (a + 1)^{2} < 90 (a + 1)^{2} .

For every integer $B \geq 1$ , either $B X_{a}$ is integral at some index (and hence at every later index), or $dist (B X_{a}, Z) \geq 1 / 31$ at arbitrarily large indices.

Proof. We count at most $(a + 1)^{2}$ smooth integers in $[2^{a}, 2^{a + 1})$ : each pair of exponents of $3, 5$ , both at most $a$ , allows at most one exponent of $2$ . By (12), $s_{a} \leq 30 (a + 1)^{2} / 8^{a}$ . As $h_{a} \leq 8^{a} / 2$ and $a + j + 1 \leq (a + 1) (j + 1)$ , we have

X_{a} \leq 15 (a + 1)^{2} \sum_{j \geq 0} \frac{(j + 1)^{2}}{8^{j}} = \frac{8640}{343} (a + 1)^{2} .

This also proves convergence. Splitting off $s_{a}$ gives the recurrence, since $h_{a + 1} = b_{a} h_{a}$ and $m_{a} = h_{a + 1} s_{a}$ .

For the final assertion, suppose that every sufficiently late distance is strictly less than $1 / 31$ , and write $B X_{a} = z_{a} + e_{a}$ , where $z_{a}$ is integral and $| e_{a} | < 1 / 31$ . The recurrence makes $e_{a + 1} - b_{a} e_{a}$ an integer of absolute value less than $(1 + b_{a}) / 31 \leq 1$ . Thus $e_{a + 1} = b_{a} e_{a}$ . A nonzero error would therefore at least double at each step, contradicting $| e_{a} | < 1 / 31$ . All sufficiently late errors vanish; integrality then persists by the recurrence. ◻

The loss of bounded tails is genuine. For $a \geq 1$ , every odd part $3^{j} 5^{k} < 2^{a}$ has a unique multiple by a power of $2$ in $[2^{a}, 2^{a + 1})$ . Taking $0 \leq j, k \leq ⌊ a / 5 ⌋$ ensures this inequality because $15 < 2^{5}$ . Each such integer contributes at least $1$ to $m_{a}$ , so

m_{a} \geq (⌊ a / 5 ⌋ + 1)^{2}, X_{a} > m_{a} / 30.

Rationality would still put the tails in a fixed lattice, but not in a finite set. Repetition alone also gives no recovery rule: the maps $y \mapsto (m_{a} + y) / b_{a}$ now depend on $a$ and form an infinite family. The proof in Section 2 needed both finiteness and unique recovery. We instead use the quadratic bound to limit the positive integer tails required by rationality.

The Cantor-series form $S / 2 = \sum_{a \geq 0} m_{a} / (b_{0} \dots b_{a})$ also explains why the criteria in [9] and [10] do not apply: their small-numerator hypothesis $m_{a} / (b_{a - 1} b_{a}) \to 0$ fails with bounded $b_{a}$ and these unbounded $m_{a}$ . The companion’s tail estimates give sharper bounds.

Theorem 3.3 (rationality gives positive integer tails). If $S = A / D$ in lowest terms, where $D = 2^{u} 3^{v} 5^{w} B$ and $gcd (B, 30) = 1$ , then for every $a \geq a_{0} = u + 1 + 2 v + 3 w$ ,

d_{a} = B X_{a} \in Z_{> 0}, d_{a + 1} = b_{a} d_{a} - B m_{a}, d_{a} \leq 90 B (a + 1)^{2} .

Proof. For $a \geq 1$ , clearing the finite prefix gives

\begin{matrix} (5) & X_{a} = h_{a} S - \sum_{\begin{matrix} x \in S_{P} \\ x < 2^{a} \end{matrix}} \frac{h_{a}}{H (x)}, \sum_{\begin{matrix} x \in S_{P} \\ x < 2^{a} \end{matrix}} \frac{h_{a}}{H (x)} \in Z . \end{matrix}

If $a \geq u + 1 + 2 v + 3 w$ , the exponent $a - 1$ of $2$ in $h_{a}$ is at least $u$ , and $2^{a} \geq 3^{v}, 5^{w}$ . Therefore $2^{u} 3^{v} 5^{w} ∣ h_{a}$ , and $B X_{a}$ differs from $h_{a} A / (2^{u} 3^{v} 5^{w})$ by an integer. Positivity, recurrence and the bound follow from Proposition 3.2. ◻

Corollary 3.4 (the first index at which the denominator clears). Under the same lowest-terms hypothesis, put $M = 2^{u} 3^{v} 5^{w}$ and let $den$ denote the positive reduced denominator. For $a \geq 1$ ,

den (B X_{a}) = \frac{M}{gcd (M, h_{a})} .

Consequently $B X_{a}$ is integral exactly when $2^{a} \geq max (2^{u + 1}, 3^{v}, 5^{w})$ . The first such $a$ can be found by integer comparisons, without logarithmic rounding.

Proof. Equation (5) shows that $B X_{a}$ and $h_{a} A / M$ have the same reduced denominator. Since $gcd (A, M) = 1$ , this denominator is $M / gcd (M, h_{a})$ . Divisibility by each of $2^{u}, 3^{v}, 5^{w}$ gives the three stated inequalities. ◻

The two denominator parts have different roles: $M$ determines when clearing occurs; $B$ is the multiplier left after clearing. For $D = 2^{3} 3^{2} 5 \cdot 7$ , the first integral $7 X_{a}$ would occur at $a = 4$ , not the sufficient index $11$ in the theorem. Thus an early nonintegral tail need not stay nonintegral. The companion’s denominator calculation also gives the reduced denominator of $X_{a}$ itself.

A residue criterion

A finite segment bounds the starting tail. Its integrality would constrain the endpoint tail to a residue class. The residue inequality expresses the same obstruction in either form.

For two steps the recurrence reads $X_{ℓ + 2} = b_{ℓ + 1} b_{ℓ} X_{ℓ} - (b_{ℓ + 1} m_{ℓ} + m_{ℓ + 1})$ . To keep track of the multiplicative and additive terms over $h$ steps, write

W_{ℓ, 0} = 1, F_{ℓ, 0} = 0, W_{ℓ, h + 1} = b_{ℓ + h} W_{ℓ, h}, F_{ℓ, h + 1} = b_{ℓ + h} F_{ℓ, h} + m_{ℓ + h} .

Induction yields

\begin{matrix} (6) & X_{ℓ + h} = W_{ℓ, h} X_{ℓ} - F_{ℓ, h} . \end{matrix}

Since $b_{a} = h_{a + 1} / h_{a}$ , its product telescopes to $W_{ℓ, h} = h_{ℓ + h} / h_{ℓ}$ . Thus

\frac{F_{ℓ, h}}{W_{ℓ, h}} = h_{ℓ} \sum_{a = ℓ}^{ℓ + h - 1} s_{a}, \frac{X_{ℓ + h}}{W_{ℓ, h}} = h_{ℓ} T_{ℓ + h} .

These finite and remaining contributions sum to $X_{ℓ}$ , both in the starting normalisation $h_{ℓ}$ , by (6). Throughout this section let

\begin{matrix} (7) & K (B, a) = 90 B (a + 1)^{2} . \end{matrix}

The bound $0 < B X_{ℓ + h} \leq K (B, ℓ + h)$ and (6) enclose the starting tail in the interval

\frac{B F_{ℓ, h}}{W_{ℓ, h}} < B X_{ℓ} \leq \frac{B F_{ℓ, h} + K (B, ℓ + h)}{W_{ℓ, h}} .

Width below $1$ gives at most one integer, not necessarily none. With $W = W_{ℓ, h}$ and $F = F_{ℓ, h}$ , the interval contains no integer exactly when

B F + K (B, ℓ + h) < W (⌊ B F / W ⌋ + 1) .

The right side is $W$ times the first integer strictly above $B F / W$ . Its excess over $B F$ is the least positive residue of $- B F$ modulo $W$ : the residue measures the gap to that integer in units of $1 / W$ .

For integers $W \geq 1, t$ , define ${lpr}_{W} (t) = 1 + ((t - 1) mod W)$ , with remainder in ${0, \dots, W - 1}$ . Its values lie in ${1, \dots, W}$ , so the zero class is represented by $W$ . This matches the open left endpoint: if $W ∣ B F$ , the next possible integer is one full unit above $B F / W$ , not $B F / W$ itself.

Lemma 4.1 (least positive residues). If $d$ is a positive integer with $d \leq K$ and $d \equiv - B F (\mod W)$ , where $W \geq 1$ , then ${lpr}_{W} (- B F) \leq K$ .

Proof. The least positive representative is at most every positive representative of its class, and in particular at most $d$ . ◻

In (6), an integral $B X_{ℓ}$ makes $B X_{ℓ + h}$ such a positive representative.

For $ℓ = 1, h = 6$ , direct iteration gives $W_{1, 6} = 648000$ and $F_{1, 6} = 524431$ . Since $K (1, 7) = 5760$ , the interval is

\frac{524431}{648000} < X_{1} \leq \frac{530191}{648000} < 1.

Equivalently, ${lpr}_{W_{1, 6}} (- F_{1, 6}) = 123569 > 5760$ certifies $X_{1} \notin Z$ , treating only $B = 1$ at $ℓ = 1$ . For denominators $M B$ , varying $B$ covers the part coprime to $30$ ; arbitrarily late starts allow each supported part $M$ to clear.

Theorem 4.2 (a residue criterion for irrationality). The number $S$ is irrational if and only if

\begin{matrix} (8) & \begin{matrix} for every B \geq 1 with gcd (B, 30) = 1 and every a_{0} \geq 1, \\ there are ℓ \geq a_{0}, h \geq 1 such that {lpr}_{W_{ℓ, h}} (- B F_{ℓ, h}) > K (B, ℓ + h) . \end{matrix} \end{matrix}

Proof. Suppose that (8) holds and $S$ is rational. By Theorem 3.3, $d_{a} = B X_{a}$ is a positive integer for every sufficiently large $a$ . Choose a segment beyond that onset. Equation (6) gives $d_{ℓ + h} \equiv - B F_{ℓ, h} (\mod W_{ℓ, h})$ , contrary to Lemma 4.1 and (8).

Conversely, suppose that $S$ is irrational and fix $B, ℓ \geq 1$ . Equation (5) makes $B X_{ℓ}$ nonintegral. The distance upwards to the next integer, $δ = ⌈ B X_{ℓ} ⌉ - B X_{ℓ} \in (0, 1)$ , is now fixed as the segment length $h$ grows. We have $W_{ℓ, h} \geq 2^{h}$ , whereas $X_{ℓ + h}$ has only quadratic growth. Consequently $δ + B X_{ℓ + h} / W_{ℓ, h}$ lies in $(0, 1)$ for sufficiently large $h$ . This ensures that the following representative lies strictly between $0$ and the modulus, so it is the least positive one:

{lpr}_{W_{ℓ, h}} (- B F_{ℓ, h}) = ⌈ B X_{ℓ} ⌉ W_{ℓ, h} - B F_{ℓ, h} = δ W_{ℓ, h} + B X_{ℓ + h} .

The positive constant $δ$ multiplies an exponentially growing $W_{ℓ, h}$ , so the first term eventually exceeds $K (B, ℓ + h)$ . This proves the inequality from every fixed start, and hence after every prescribed onset $a_{0}$ . No bound uniform in $B$ or $ℓ$ is needed for this direction. ◻

At a fixed start, success rules out integrality of $B X_{ℓ}$ . If $B X_{ℓ}$ is integral, the recurrence, positivity and the bound prevent success at any length; for all sufficiently large $h$ , ${lpr}_{W_{ℓ, h}} (- B F_{ℓ, h}) = B X_{ℓ + h}$ . If $B X_{ℓ}$ is nonintegral, every sufficiently long segment succeeds, even when $S$ is rational with a denominator not yet cleared at $ℓ$ . Increasing the length does not remove the need for starts beyond every cutoff. A finite failed search distinguishes neither case.

The converse obtains $δ > 0$ from irrationality. Proving (8) without that assumption is the remaining task. The upper-bound property of $K$ is essential: a zero bound would make every residue inequality automatic. The companion’s residue analysis treats the fixed-start formula and smaller or more general bounds.

The repeated three-prime question can therefore be written in three equivalent forms.

Problem 4.3 (the repeated three-prime target). Prove that $S = R_{{2, 3, 5}}$ is irrational.

Problem 4.4 (no reduced tail is an integer). Prove that for every $a \geq 1$ and every integer $B \geq 1$ coprime to $30$ ,

\begin{matrix} (9) & B X_{a} \notin Z . \end{matrix}

Problem 4.5 (residue inequalities after every starting index). Prove (8).

The equivalence follows from (5), denominator clearing and Theorem 4.2; none of the three assertions is proved for $S$ . The companion’s discussion of the repeated sum records finite calculations and the hypotheses missing from proposed finite-difference and function-theoretic arguments.

Two-prime transcendence

For two primes, the distinct sum is affine in one Hecke–Mahler value; the repeated sum is quadratic in it. Fan gave the factorisation of the repeated sum and its transcendence consequence in his post of 26 June 2026 [7].

Theorem 5.1 (both two-prime sums). Let $p < q$ be distinct primes. Put $θ = \log p / \log q$ and $A = \sum_{n \geq 0} p^{- n} q^{- ⌊ n θ ⌋}$ . Let $R_{p, q}$ sum the reciprocal running LCM at every positive ${p, q}$ -smooth integer, and let $D_{p, q}$ count each distinct running LCM once. Then

\begin{matrix} (10) & D_{p, q} = \frac{(q - p) A + p}{q - 1}, R_{p, q} = \frac{(p + q - 1) A - (p - 1) A^{2}}{q - 1} . \end{matrix}

Both numbers are transcendental.

Proof. Write $x = 1 / p$ , $y = 1 / q$ , $m_{n} = ⌊ n θ ⌋$ and $δ_{n} = m_{n + 1} - m_{n}$ . Unique factorisation makes $θ$ irrational, and $0 < θ < 1$ gives $δ_{n} \in {0, 1}$ . The initial value and the positive powers of $p$ contribute $A = \sum_{n \geq 0} x^{n} y^{m_{n}}$ . There is one $q$ -power strictly between $p^{n}$ and $p^{n + 1}$ exactly when $δ_{n} = 1$ , so the positive powers of $q$ contribute

B = \sum_{n \geq 0} δ_{n} x^{n} y^{m_{n} + 1}, D_{p, q} = A + B .

Absolute convergence allows an index shift. Multiplying $y^{m_{n + 1}} - y^{m_{n}} = δ_{n} y^{m_{n}} (y - 1)$ by $x^{n + 1}$ and summing gives $(A - 1) - x A = x (y - 1) B / y$ . Therefore

B = \frac{p - (p - 1) A}{q - 1} .

For the repeated sum, the running LCM at $p^{i} q^{j}$ is $p^{i + ⌊ j / θ ⌋} q^{j + m_{i}}$ . Its reciprocal is the product of $x^{i} y^{m_{i}}$ , which depends only on $i$ , and $y^{j} x^{⌊ j / θ ⌋}$ , which depends only on $j$ . Absolute convergence therefore gives

R_{p, q} = A \sum_{j \geq 0} y^{j} x^{⌊ j / θ ⌋} = A (1 + B) .

For the last equality, $n = ⌊ j / θ ⌋$ locates $q^{j}$ between $p^{n}$ and $p^{n + 1}$ ; hence $m_{n} = j - 1$ and $δ_{n} = 1$ . This matches the positive- $j$ terms with those of $B$ , while $j = 0$ contributes $1$ . Substitution proves (10).

Factorisation alone would not prove transcendence of the product. The two identities instead express both sums as nonconstant polynomials over $Q$ in the same $A$ . It remains to prove $A$ transcendental. For this we use the Hecke–Mahler series $F_{θ} (x, y) = \sum_{n \geq 1} \sum_{k = 1}^{m_{n}} x^{n} y^{k}$ . Geometric summation gives

\begin{matrix} (11) & A = \frac{1}{1 - x} - \frac{1 - y}{y} F_{θ} (x, y) . \end{matrix}

Bugeaud and Laurent’s theorem [5] applies with intercept $ρ = 0$ , $β = x$ and $α = y$ : the slope $θ$ is irrational and lies in $(0, 1)$ , $x$ and $y$ are nonzero algebraic numbers, $| x | < 1$ and $| x y^{θ} | = p^{- 2} < 1$ . Thus $F_{θ} (x, y)$ is transcendental; this case $ρ = 0$ is due to Loxton and van der Poorten [6]. Thus $A$ is transcendental. The displayed affine and quadratic polynomials are nonconstant, so an algebraic value of either would force $A$ to be algebraic. ◻

The companion’s two-prime calculation records an ordinary extension to the monoid generated by coprime integers $1 < p < q$ , such as $4, 9$ . That monoid need not contain all integers supported on the prime divisors of $p q$ ; multiplicatively dependent generators such as $4, 8$ require a different calculation.

For larger prime sets, the sums over pure powers $E_{p} = \sum_{α \geq 0} L (p^{α})^{- 1}$ satisfy $D_{P} = \sum_{p \in P} E_{p} - (| P | - 1)$ . The companion’s prime-power subseries argument treats their irrationality as a remark with an ordinary proof awaiting formalisation. The source record contains no separate priority search for this claim. Irrationality of the individual subseries would not imply irrationality of their sum. The repeated series also contains mixed products of primes, whose multiplicities change the tail estimates.

Identities for the running LCM and multiplicities

Let $p, q, r$ be pairwise distinct primes. Write $H = H_{{p, q, r}}$ and $K (i, j, k) = H (p^{i} q^{j} r^{k})^{- 1}$ throughout this appendix and the next. The following identities underlie the grouping of the repeated series in Section 3.

Proposition A.1 (the running least common multiple). Let $p, q, r$ be pairwise distinct primes and $x \geq 1$ . Then the running least common multiple is given by $L (x) = H (x)$ .

Proof. We have $u ∣ H (x)$ for every supported $u \leq x$ . For the reverse divisibility, take the maximal powers of $p, q, r$ below $x$ : they belong to the prefix and are pairwise coprime, so their product divides its LCM. ◻

For each prime, $x / p < p^{⌊ \log_{p} x ⌋} \leq x$ . Multiplying gives

\begin{matrix} (12) & x^{3} / (p q r) < H (x) \leq x^{3} . \end{matrix}

Proposition A.2 (cells and jumps). The running LCM is constant when the three integer logarithms are constant. A jump in exactly one logarithm multiplies it by the corresponding prime. The first $n$ positive powers of each prime, together with $1$ , form $3 n + 1$ distinct points.

Proof. The running LCM is a product of the three maximal pure powers, which proves the constancy and jump assertions. Unique factorisation separates the $3 n$ positive pure powers from one another and from $1$ . ◻

Proposition A.3 (grouping terms with the same running LCM). For a finite exponent box $B$ , set $F (H) = {(i, j, k) \in B : H (p^{i} q^{j} r^{k}) = H}$ . Then

\begin{matrix} (13) & \sum_{(i, j, k) \in B} K (i, j, k) = \sum_{H} \frac{# F (H)}{H} . \end{matrix}

Proof. Partition the finite exponent box by the value of the running LCM. The terms in the part $F (H)$ all equal $1 / H$ . ◻

The three-prime kernel

The two-prime proof separates the summand by exponents. For three primes, the carry in adding two fractional parts prevents any finite separation of $i$ from $(j, k)$ . Diagonal rescaling leaves just two entry values, $1$ and $1 / r$ ; choosing their threshold positions gives nonzero minors of every order. This concerns exact kernel identities, not identities after summation.

Example B.1. For $(p, q, r) = (2, 3, 5)$ , the leading two-by-two determinant is $- 1 / 15$ : its four entries are $1, 1 / 6, 1 / 2, 1 / 60$ , so it equals $1 / 60 - 1 / 12$ .

A sum of $d$ separated products has rank at most $d$ in each layer. Thus the two-by-two example excludes only rank one; excluding every finite $d$ requires nonzero minors of arbitrary order. Their indices may depend on the order, but not on $k$ .

Theorem B.2 (no finite separation of the kernel). Let $p, q, r$ be primes with $p \neq q$ , $p \neq r$ and $q \neq r$ . For every $n \geq 0$ there are injective maps $I, J : {0, \dots, n - 1} \to N$ such that, for every $k \geq 0$ ,

det (K (I (a), J (b), k))_{0 \leq a, b < n} \neq 0.

Consequently, for no finite $d$ do there exist rational-valued functions $f_{ℓ} (i)$ and $G_{ℓ} (j, k)$ , $0 \leq ℓ < d$ , satisfying

K (i, j, k) = \sum_{ℓ < d} f_{ℓ} (i) G_{ℓ} (j, k) for all i, j, k .

Proof. Fix $k$ . The powers of $p$ and $q$ separate into row and column factors, as before. Only the power of $r$ retains an interaction between $i$ and $j$ . Put $c = r^{- 1}$ , $x_{i} = {i \log_{r} p}$ and $y_{j} = {j \log_{r} q}$ , where braces denote fractional parts. Then

⌊ i \log_{r} p + j \log_{r} q ⌋ = ⌊ i \log_{r} p ⌋ + ⌊ j \log_{r} q ⌋ + 1_{{x_{i} + y_{j} \geq 1}} .

The last term is the carry when the two fractional parts are added. Putting all the other prime powers into row and column factors gives

K (i, j, k) = U_{i} (k)^{- 1} C_{i j} V_{j} (k)^{- 1}, C_{i j} = c^{1_{{x_{i} + y_{j} \geq 1}}},

where

\begin{aligned} U_{i} (k) & = p^{i} q^{⌊ \log_{q} (p^{i} r^{k}) ⌋} r^{k + ⌊ \log_{r} p^{i} ⌋}, \\ V_{j} (k) & = q^{j} p^{⌊ \log_{p} (q^{j} r^{k}) ⌋} r^{⌊ \log_{r} q^{j} ⌋} . \end{aligned}

The positive factors preserve nonvanishing of minors and contain all dependence on $k$ . In column $j$ , the entry changes from $1$ to $c$ at the threshold $1 - y_{j}$ . Both $\log_{r} p$ and $\log_{r} q$ are irrational, so their fractional-part orbits are dense separately; no joint orbit is needed. For $n \geq 1$ , choose rows with $0 < x_{I (0)} < \dots < x_{I (n - 1)} < 1$ . Then choose each column threshold between consecutive row phases. Writing $s_{b} = 1 - y_{J (b)}$ , take

s_{0} \in (0, x_{I (0)}), s_{b} \in (x_{I (b - 1)}, x_{I (b)}) (1 \leq b < n) .

The intervals for the $s_{b}$ are disjoint, so the column indices are distinct. The strict inequalities also avoid the threshold itself, and give $x_{I (a)} + y_{J (b)} \geq 1$ exactly when $b \leq a$ . Thus multiplying row $a$ by $U_{I (a)} (k)$ and column $b$ by $V_{J (b)} (k)$ reduces the selected kernel minor to

T_{n} (c) = (\begin{matrix} c & 1 & \dots & 1 \\ c & c & \dots & 1 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ c & c & \dots & c \end{matrix}), det T_{n} (c) = c (c - 1)^{n - 1} .

Indeed, subtract each preceding row from the next, working upwards from the last row. Each row difference has just one nonzero entry, $c - 1$ , in its own column. The resulting triangular matrix has diagonal entries $c, c - 1, \dots, c - 1$ . The original determinant is therefore

c (c - 1)^{n - 1} \prod_{a < n} U_{I (a)} (k)^{- 1} \prod_{b < n} V_{J (b)} (k)^{- 1} \neq 0.

The indices were chosen from $C$ , independently of $k$ . For $n = 0$ the empty determinant equals $1$ .

For the final assertion, suppose a separation with $d$ summands existed and fix any $k$ . On the rows $I (0), \dots, I (d)$ and columns $J (0), \dots, J (d)$ its matrix would be the product of the $(d + 1) \times d$ matrix $(f_{ℓ} (I (a)))_{a, ℓ}$ and the $d \times (d + 1)$ matrix $(G_{ℓ} (J (b), k))_{ℓ, b}$ . Its rank would be at most $d$ , so its determinant would vanish, contrary to the minor just constructed. ◻

No continuity or boundedness is assumed for the separated factors.

Corollary B.3 (the same minors modulo integers coprime to $30$ ). For $(p, q, r) = (2, 3, 5)$ and every $n \geq 1$ , there are injective maps $I, J : {0, \dots, n - 1} \to N$ , chosen independently of $B$ and $k$ , such that for every $B \geq 2$ coprime to $30$ and every $k \geq 0$ , the selected $n \times n$ kernel matrix has unit determinant over $Z / B Z$ , with each reciprocal prime power interpreted by its modular inverse.

Proof. Choose the maps from Theorem B.2. Every row and column factor is a unit modulo $B$ . The normalised determinant is $5^{- 1} (- 4 / 5)^{n - 1}$ , also a unit. ◻

The modulus may be composite, for example $49$ or $77$ . Coprimality with $30$ makes both the reciprocal entries and the determinant units: the latter introduces only a power of $4$ in its numerator.

Leading minors.

The leading $4 \times 4$ block at ${2, 3, 5}$ is singular:

K (3, j, 0) = \frac{1}{120} K (0, j, 0) (0 \leq j < 4) .

Evaluate the running LCM at $3^{j}$ and $8 \cdot 3^{j}$ to obtain these equalities. At $j = 4$ , however, $K (3, 4, 0) - K (0, 4, 0) / 120 = - 1 / 19440000$ . The selection in the theorem avoids this problem by choosing column thresholds at distinct cuts among the row phases. An arbitrary sample may miss gaps; its rank counts distinct cuts, with one exception.

Proposition B.4 (rank of a matrix of threshold columns). Let $m \geq 1$ and let $c$ lie in a field with $c \neq 0, 1$ . For $0 \leq k \leq m$ write $v_{k}$ for the length- $m$ column with a $1$ in each of the first $k$ coordinates and $c$ thereafter. A matrix whose distinct columns are $v_{k}$ for $k$ in a nonempty set $E$ has rank $| E | - 1_{{0, m} \subseteq E}$ .

The exception is the pair of constant columns, $v_{0} = c v_{m}$ . All other distinct cuts contribute independently.

Proof. List $E = {k_{1} < \dots < k_{t}}$ . The $t - 1$ differences $v_{k_{j + 1}} - v_{k_{j}} = (1 - c) 1_{{k_{j}, \dots, k_{j + 1} - 1}}$ have disjoint nonempty supports, hence are linearly independent. If $k_{1} > 0$ or $k_{t} < m$ , their union misses a coordinate where $v_{k_{1}}$ is nonzero, so the rank is $t$ . If $k_{1} = 0$ and $k_{t} = m$ , then $v_{0} = c 1$ lies in the span of the differences (their sum is $(1 - c) 1$ ), so the rank is $t - 1$ . ◻

Order the sampled row phases and take $c = 1 / r$ over $Q$ . The formula then gives the rank of every nonempty rectangular sample, independently of $k$ : the layer changes only the nonzero diagonal factors. The companion’s kernel analysis gives the exact rational comparisons.

Approximation.

The diagonal rescaling preserves rank but removes the kernel’s decay. Retaining rows $i < N$ in the original kernel gives $N$ separated terms; $H (x) > x^{3} / (p q r)$ makes the error tend to zero uniformly and in $ℓ^{1} (N^{3})$ . The rescaled matrix instead has uniform distance $(1 - 1 / r) / 2$ from finite-rank matrices. The companion’s approximation theorem proves both statements. Neither implies irrationality of the scalar sum.

Verification, reproducibility and sources

The five-map proof is in the distinct-value development and its block-series criterion. Its statement has no recorded Comparator comparison. The evidence record fixes the source versions; the named input for two-prime transcendence is stated beside that theorem.

The programs for sums over distinct values calculate the interval endpoints and finite-state examples using exact arithmetic; the proof in Section 2 derives its constants directly.

The companion compares one-jump and dyadic bounds. Its finite-state extension uses earlier blocks to sharpen the intervals. Saved exact computations satisfy the criterion for every three-prime set and $292$ of the $330$ four-prime sets of primes at most $31$ . The criterion and computations remain remarks awaiting formalisation. For ${2, 3, 5, 7}$ , two admissible automaton words induce the same affine map, defeating first-block recovery at every refinement. Equal maps alone do not give equal arithmetic tails.

The companion’s source inventory covers the ordinary general and subseries arguments, the 26 September 2026 AI check and its corrections. No independent human mathematical review has been reported.

Further comparisons

The companion compares our bounded-base series with the criteria of Erdős–Straus [9], Hančl–Tijdeman [10] and Diananda–Oppenheim [8]; the bounded-ratio argument with Erdős–Taylor [11] and Fan [12]; and integer recurrences with Hančl–Tijdeman’s polynomial Cantor-series results [13]. Koutsoukou-Argyraki and Li’s Isabelle development [19] formalises classical criteria, not this repeated series.

The transcendence comparisons concern finite differences [14], echoing [15], fixed-base digit complexity [20] and multivariate Mahler theory [21]. The companion exhibits nonzero third-difference boundary terms and distinguishes variable-base digits from fixed-base expansions: quadratic growth supplies neither the missing cancellation nor recurrence hypotheses. Fixed-prime semigroups [16][17] and other Ahmes-series problems [18] supply context, not estimates used in our counting proof.

Statements and declarations

Authorship and availability.

AI agents carried out most of the research and drafting. The companion contains the longer proofs and exact computations, together with the hypotheses and limitations of the unsuccessful approaches.

Funding and competing interests.

This work received no external funding. The author declares no competing interests.

Acknowledgements.

The problem number and its status as of 28 July 2026 are taken from Thomas Bloom’s Erdős Problems catalogue [4].

References

Paul Erdős and Ronald L. Graham, Old and New Problems and Results in Combinatorial Number Theory, Monographies de L’Enseignement Mathématique 28, L’Enseignement Mathématique (1980), source.
Paul Erdős, On the irrationality of certain series: problems and results, in New Advances in Transcendence Theory, Cambridge University Press (1988), 102–109, doi:10.1017/CBO9780511897184.009.
Paul Erdős, Letter to the Editor, Fibonacci Quarterly 12, no. 4 (1974), 335, source.
Thomas F. Bloom, Erdős Problem #269 (2026), source. Accessed 28 July 2026.
Yann Bugeaud and Michel Laurent, Transcendence and continued fraction expansion of values of Hecke–Mahler series, Acta Arithmetica 209 (2023), 59–90, doi:10.4064/aa220323-18-1; arXiv:2203.12901.
John H. Loxton and Alfred J. van der Poorten, Arithmetic properties of certain functions in several variables III, Bulletin of the Australian Mathematical Society 16 (1977), 15–47, doi:10.1017/S0004972700022978.
Steve Fan, Comment on Erdős Problem #269, thread 269, post 7218 (2026), source. Public forum post, 26 June 2026, thread 269, post 7218.
P. H. Diananda and A. Oppenheim, Criteria for irrationality of certain classes of numbers II, American Mathematical Monthly 62, no. 4 (1955), 222–225. Cited in the form stated by Serbenyuk, arXiv:1706.03124, Theorem 1.
Paul Erdős and Ernst G. Straus, On the irrationality of certain series, Pacific Journal of Mathematics 55, no. 1 (1974), 85–92, doi:10.2140/pjm.1974.55.85.
Jaroslav Hančl and Robert Tijdeman, On the irrationality of Cantor and Ahmes series, Publicationes Mathematicae Debrecen 65, no. 3–4 (2004), 371–380, doi:10.5486/PMD.2004.3254.
Paul Erdős and S. James Taylor, On the set of points of convergence of a lacunary trigonometric series and the equidistribution properties of related sequences, Proceedings of the London Mathematical Society s3-7, no. 1 (1957), 598–615, doi:10.1112/plms/s3-7.1.598.
Steve Fan, Strongly complete sets and a conjecture of Erdős (2026), source; arXiv:2607.14071. The cited Lemma 3.1 is in arXiv v1, 15 July 2026.
Jaroslav Hančl and Robert Tijdeman, On the irrationality of polynomial Cantor series, Acta Arithmetica 133, no. 1 (2008), 37–52, doi:10.4064/aa133-1-3.
Florian Luca, Joël Ouaknine and James Worrell, Transcendence of Hecke–Mahler Series, Bulletin of the London Mathematical Society 57, no. 5 (2025), 1360–1368, doi:10.1112/blms.70033; arXiv:2412.07908. Numbered references use the published article.
Pavol Kebis, Florian Luca, Joël Ouaknine, Andrew Scoones and James Worrell, On Transcendence of Numbers Related to Sturmian and Arnoux-Rauzy Words, in 51st International Colloquium on Automata, Languages, and Programming (ICALP 2024), Leibniz International Proceedings in Informatics 297, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2024), 144:1–144:15, doi:10.4230/LIPIcs.ICALP.2024.144.
Robert Tijdeman and H. G. Meijer, On integers generated by a finite number of fixed primes, Compositio Mathematica 29, no. 3 (1974), 273–286, source.
Alessandro Languasco, Florian Luca, Pieter Moree and Alain Togbé, Sequences of integers generated by two fixed primes, Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 95 (2025), 123–148, doi:10.1007/s12188-025-00293-9; arXiv:2309.12806.
Vjekoslav Kovač and Terence Tao, On several irrationality problems for Ahmes series, Acta Mathematica Hungarica 175 (2025), 572–608, doi:10.1007/s10474-025-01528-0; arXiv:2406.17593.
Angeliki Koutsoukou-Argyraki and Wenda Li, Irrationality Criteria for Series by Erdős and Straus, Archive of Formal Proofs (2020), source. Entry dated 12 May 2020; proof-document version consulted: 6 February 2026.
Boris Adamczewski and Yann Bugeaud, On the complexity of algebraic numbers I. Expansions in integer bases, Annals of Mathematics 165, no. 2 (2007), 547–565, doi:10.4007/annals.2007.165.547.
Boris Adamczewski and Colin Faverjon, Mahler’s method in several variables and finite automata, Annals of Mathematics 204, no. 2 (2026), 455–533, doi:10.4007/annals.2026.204.2.1. Online 13 September 2026; locators here refer to the 68-page author manuscript.

Long paper, 73 pages

Running least common multiples: distinct values and multiplicities

PDFOpen the PDF, 73 pages Open as its own page Current source

PDF and source links follow the latest files on GitHub. The web reader shows the edition published here.

Lean source files

71 source files: 70 programme or result files and 1 imported dependency. See them in the Lean map.

The programme directory, linked result sources and their local imports. Shared files can appear in more than one programme.

GitHub lean/ErdosProblems/Erdos269/ActualSharpTailMajorantR10.lean
GitHub lean/ErdosProblems/Erdos269/BlockMassEngines.lean
GitHub lean/ErdosProblems/Erdos269/BoundedRadixTailEscape.lean
GitHub lean/ErdosProblems/Erdos269/CarryLiftExtinction.lean
GitHub lean/ErdosProblems/Erdos269/CertificateRangeR11.lean
GitHub lean/ErdosProblems/Erdos269/CertificateSafetyR12.lean
GitHub lean/ErdosProblems/Erdos269/CofinalWindowEscapeEquivalence.lean
GitHub lean/ErdosProblems/Erdos269/DistinctHeightAxiomAudit.lean
GitHub lean/ErdosProblems/Erdos269/DistinctHeightBlockRadix.lean
GitHub lean/ErdosProblems/Erdos269/DistinctHeightIrrationality.lean
GitHub lean/ErdosProblems/Erdos269/DyadicBlockMassIdentity.lean
GitHub lean/ErdosProblems/Erdos269/DyadicBlockThresholdPartition.lean
GitHub lean/ErdosProblems/Erdos269/DyadicOrderedTailRecurrence.lean
GitHub lean/ErdosProblems/Erdos269/DyadicRadixTailEscape.lean
GitHub lean/ErdosProblems/Erdos269/DyadicShellSummability.lean
GitHub lean/ErdosProblems/Erdos269/FiniteCutRank.lean
GitHub lean/ErdosProblems/Erdos269/IntegralBranchExtinction.lean
GitHub lean/ErdosProblems/Erdos269/IntegralBranchWidth.lean
GitHub lean/ErdosProblems/Erdos269/JumpConstraintMajorant.lean
GitHub lean/ErdosProblems/Erdos269/KernelCarryRank.lean
GitHub lean/ErdosProblems/Erdos269/LongWindowCapR11.lean
GitHub lean/ErdosProblems/Erdos269/NormalizedStateWidth.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/BoundedLatticeCollision.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/DyadicAlphabetWhole.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/EightScaleRigidity.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/FixedBaseRecoding.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/LiteralTriangle.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/LiteralTriangleReal.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/OcticEscapeWhole.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/PhaseStripDecomposition.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/RealBaseShortInterval.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/RealBoundDenominatorReduction.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/RealCutoffs.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/RealTwoPrimeKernel.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/StripDecomposition.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/UniformRankAudit.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/WeightedShiftArithmetic.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR20/WeightedShiftValue.lean
GitHub lean/ErdosProblems/Erdos269/PaperCompleteR21/TwoPrimeSums.lean
GitHub lean/ErdosProblems/Erdos269/PaperCorrespondenceAudit.lean
GitHub lean/ErdosProblems/Erdos269/PaperExactDenominatorR13.lean
GitHub lean/ErdosProblems/Erdos269/PaperFixedStartResidueR14.lean
GitHub lean/ErdosProblems/Erdos269/PaperR12WaveAAxiomAudit.lean
GitHub lean/ErdosProblems/Erdos269/PaperR14StatementAxiomAudit.lean
GitHub lean/ErdosProblems/Erdos269/PaperR7ActualOrbit.lean
GitHub lean/ErdosProblems/Erdos269/PaperR7AnalyticInterfaces.lean
GitHub lean/ErdosProblems/Erdos269/PaperR7BasicAssembly.lean
GitHub lean/ErdosProblems/Erdos269/PaperR7FiniteCutRank.lean
GitHub lean/ErdosProblems/Erdos269/PaperR7ModularMinors.lean
GitHub lean/ErdosProblems/Erdos269/PaperR7RationalBridge.lean
GitHub lean/ErdosProblems/Erdos269/PaperR7SeriesIdentification.lean
GitHub lean/ErdosProblems/Erdos269/PaperR7SharpShellBound.lean
GitHub lean/ErdosProblems/Erdos269/PaperR7WindowResults.lean
GitHub lean/ErdosProblems/Erdos269/PaperR8RankMajorant.lean
GitHub lean/ErdosProblems/Erdos269/PaperR8UniformRank.lean
GitHub lean/ErdosProblems/Erdos269/PaperR9SourceCounts.lean
GitHub lean/ErdosProblems/Erdos269/PaperR9WidthData.lean
GitHub lean/ErdosProblems/Erdos269/PurePowerIrrationality.lean
GitHub lean/ErdosProblems/Erdos269/R12/ExactModulus.lean
GitHub lean/ErdosProblems/Erdos269/R12/OcticWindowBand.lean
GitHub lean/ErdosProblems/Erdos269/R12/PrefixExtension.lean
GitHub lean/ErdosProblems/Erdos269/RationalLatticeReduction.lean
GitHub lean/ErdosProblems/Erdos269/RationalityCarryBridge.lean
GitHub lean/ErdosProblems/Erdos269/RealCutoffR10.lean
GitHub lean/ErdosProblems/Erdos269/ResidueEscape.lean
GitHub lean/ErdosProblems/Erdos269/RestrictedFloorSum.lean
GitHub lean/ErdosProblems/Erdos269/SharpWindowCapR10.lean
GitHub lean/ErdosProblems/Erdos269/ThreeChannelBlockRigidity.lean
GitHub lean/ErdosProblems/Erdos269/ThreePrimeRunningLcm.lean
GitHub lean/ErdosProblems/Erdos269/WeightedPhaseCarry.lean
GitHub lean/ErdosProblems/Shared/IrrationalRotationStaircase.lean

Statements with Lean proofs

Lean is software that verifies a formal proof against its theorem statement and assumptions. These proofs establish only the statements listed here.

That for three pairwise distinct primes the least common multiple of the smooth prefix equals the product of the three maximal pure prime powers below the cutoff.
That equal logarithmic coordinates give an equal prefix least common multiple, so the running value is constant on logarithmic cells.
That advancing exactly one logarithmic coordinate multiplies the running value by the corresponding prime.
That the positive power channels of distinct primes are disjoint, so the first powers in three channels give an exact jump count.
That a finite smooth lattice sum groups exactly by running height, with the height fibre cardinality as coefficient.
A quadratic bound on the cardinality of the actual filtered exponent shell.
That each dyadic block contains at most one internal pure power from each non-dyadic channel, so the compressed {2,3,5} block radix belongs to the exact alphabet {2,6,10,30} and in particular lies between 2 and 30.
That the smallest {2,3,5} kernel rectangle has determinant exactly $- 1 / 15$ , so its two-by-two matrix has rank two over the rationals and no rank-one separation exists.
The canonical least-positive-residue interval and congruence facts, including the zero-residue representative.
The exact local-window identity for an integral carry.
For the actual {2,3,5} series, rationality produces a positive reduced carry; cofinal local-window escape rules out such carries under the recorded bound.
A bounded-radix dichotomy: under a bounded positive digit alphabet, a nonzero scaled tail must be separated from the integers cofinally.
Exact three-channel block-nullity as a coboundary condition, and extinction of a bounded carry lift under two independent anchors.
The exact residue-digit plus coboundary form of an integral carry, together with a function-faithful observer criterion and a finite realised-span consequence.
Irrationality of the distinct-height sum for {2,3,5}, which counts each running value once.

What these proofs do not establish

Irrationality of the repeated sum, or transcendence, in any three-prime case.
CofinalLocalWindowEscape for those actual words; it is an explicit hypothesis of the checked consumer, not a theorem.
A proof that the actual three-channel word satisfies the block-null, anchor, or factorisation hypotheses of the migrated abstract consumers.
The infinite limit of the jump expansion; only its finite core is checked.
Any unbounded denominator exclusion. The finite certificates exclude denominators below an explicit bound and no further.

The formal record: 3 main results, 3 open obligations, 7 selected result modules, finite search evidence, Comparator review

Main results

Statements and their limits are copied from the public research record. Recorded status: registered_declarations

no-go result family: rank_two_kernel_no_go

The smallest displayed 2,3,5 kernel minor is exactly $- 1 / 15$ .

Failure of rank one does not imply irrationality.

ErdosProblems.Erdos269.kernel_235_minor_eq_neg_one_fifteen

locally proved result; novelty unassessed family: three_prime_lcm_cells

The running lcm of the three-prime smooth prefix equals the exact coordinatewise height.

The identity does not supply the missing irrationality bridge.

ErdosProblems.Erdos269.smoothPrefixLcm_eq_threePrimeHeight

conditional no-go consumer; novelty and significance unassessed family: conditional_carry_escape

Under the denominator-dependent cofinal local-window residue-escape predicate, no positive reduced carry can satisfy the exact multiplier recurrence together with its short bound.

The representative assumes b,m : ℕ → ℕ, shortBound : ℕ → ℕ → ℕ, CofinalLocalWindowEscape b m shortBound, and for each positive B coprime to 30 a positive integer-valued d with d(n+1) = (b n : ℤ) d(n) − (B : ℤ) (m n : ℤ) and Int.natAbs (d n) ≤ shortBound B n; it then gives False. The representative remains a conditional consumer for arbitrary words. For the actual three-prime word, the rationality-to-reduced-carry bridge is established separately; the source-specific cofinal escape producer is not proved. This is not a #269 endpoint or irrationality proof, and no actual-series identification, novelty, priority, significance, or external-review claim is made. It is one conditional window-carry family, distinct from the finite residue, rank-two, and weighted-phase observer families.

ErdosProblems.Erdos269.no_positive_reducedCarry_of_cofinalLocalWindowEscape

Assumptions still needing proofs

Prove any one of these and the corresponding reduction above becomes unconditional.

actual_local_window_residue_escape For every positive B coprime to 30 and every starting index, produce a later actual {2,3,5} window whose least positive forcing residue exceeds the exact denominator-dependent carry bound.
nonstationary_analytic_theorem Prove a genuinely higher-dimensional analytic theorem for the phase cocycle or the associated contraction.
unbounded_height_certificate Replace one finite denominator exclusion by a family whose exclusion bound tends to infinity.

Finite search evidence

The integer-only dyadic-window checker reproduces three exact packet certificates and, for all 106,666 pairs with denominator $B \leq 1000$ coprime to 30 and dyadic start $100 \leq a \leq 500$ , finds an escaping window of length at most 18; the largest first successful length is 14. This is finite proof-consumer evidence only and proves neither unbounded-denominator nor cofinal escape.

Selected Lean proof files

The public result record selects 7 modules. Browse all 71 source files in this programme’s source map, including shared result sources and local imported dependencies.

Module	Lines	Declarations	Theorems
`ErdosProblems.Erdos269.ThreePrimeRunningLcm`	912	55	39
`ErdosProblems.Erdos269.ResidueEscape`	164	8	6
`ErdosProblems.Erdos269.RestrictedFloorSum`	709	43	30
`ErdosProblems.Erdos269.BoundedRadixTailEscape`	304	10	8
`ErdosProblems.Erdos269.CarryLiftExtinction`	351	14	12
`ErdosProblems.Erdos269.ThreeChannelBlockRigidity`	128	6	4
`ErdosProblems.Erdos269.WeightedPhaseCarry`	371	23	14

Comparator review

Comparator runs selected Lean proofs against separately supplied theorem statements using recorded software versions. A successful run establishes that those proofs satisfy those statements; it does not prove the unanswered questions above.

Replayed by the Comparator

Statement and type parity replayed against the pinned toolchain.

Exact running-LCM product, logarithmic-cell constancy, coordinate jump ratios, and jump count.

Cell structure alone does not prove irrationality.

three_prime_lcm_cells: targeted_running_lcm_identity
The 2,3,5 kernel is not rank one and its smallest displayed minor equals $- 1 / 15$ .

Failure of rank one does not itself imply irrationality.

rank_two_kernel_no_go: targeted
Under the denominator-dependent cofinal local-window residue-escape predicate, no positive reduced carry can satisfy the exact multiplier recurrence together with its short bound. The load-bearing consumer is no_positive_reducedCarry_of_cofinalLocalWindowEscape; an absorbed nonzero common-factor carry reduces exactly to that consumer. CofinalLocalWindowEscape, windowBase, windowForcing, leastPositiveResidue, and the absorbed-carry bridge are subordinate finite-window mechanism evidence.

The representative assumes b,m : ℕ → ℕ, shortBound : ℕ → ℕ → ℕ, CofinalLocalWindowEscape b m shortBound, and for each positive B coprime to 30 a positive integer-valued d with d(n+1) = (b n : ℤ) d(n) − (B : ℤ) (m n : ℤ) and Int.natAbs (d n) ≤ shortBound B n; it then gives False. The absorbed-carry bridge additionally assumes smoothFactor ≠ 0, c n = smoothFactor * d n, and the corresponding exact absorbed recurrence, then cancels that factor to the same reduced consumer. The representative remains a conditional consumer for arbitrary words. For the actual three-prime word, the rationality-to-reduced-carry bridge is established separately; the source-specific cofinal escape producer is not proved. This is not a #269 endpoint or irrationality proof, and no actual-series identification, novelty, priority, significance, or external-review claim is made. It is one conditional window-carry family, distinct from the finite residue, rank-two, and weighted-phase observer families.

conditional_carry_escape: targeted
An exact weighted-phase carry recurrence splits into a finite residue digit and an uncontrolled integral quotient coboundary; an explicit function-faithful finite-dimensional observer then forces finite realised span.

The recurrence supplies a finite residue coordinate but leaves an uncontrolled integral quotient coboundary; it proves neither a finite-state quotient nor a literal infinite realised span. Finite realised span requires an explicit factorisation through a finite-dimensional function-faithful observer, and scalar evaluation alone is insufficient. No irrationality conclusion follows from this observer alone. The actual rationality-to-positive-reduced-carry bridge is established separately; cofinal local-window escape remains open.

weighted_phase_carry_observer: targeted

Represented indirectly

Covered through a neighbouring selected interface, not replayed one-to-one.

Finite height-fibre normal form and a quadratic smooth-shell multiplicity bound.

These fibre bounds are counting input, not an escape producer. The separate actual-series divisibility bridge is established.

height_fibre_and_shell: represented_by_three_prime_structure_target
The exact dyadic block alphabet is 2, 6, 10, and 30.

The finite alphabet does not supply the needed carry escape.

dyadic_block_alphabet: represented_by_three_prime_structure_target

Outside the Comparator selection

Stated in the corpus; not part of the replayed interface set.

Steve Fan posted the two-prime factorisation, reduction and transcendence first (erdosproblems.com/269, 26 June 2026, post-7218). This checkout records that ordinary deduction from Bugeaud–Laurent and does not claim priority or a Lean theorem.

Comparator cannot certify the external analytic input.

two_prime_transcendence: not_applicable_not_a_lean_declaration
A local-window checker searches 106666 denominator and start pairs and records small certificates.

Finite search is not a cofinal statement.

three_prime_finite_search: not_applicable_not_a_lean_theorem
For positive rationally independent clock rates with no simultaneous positive crossings, and integer radices $b_{i} \geq 2$ , all rational relations among 1 and the channel sums $S_{i}$ are multiples of $(- 1, b_{1} - 1, \dots, b_{d} - 1)$ . The proof uses genuine late swaps with a common infinite suffix in one orbit closure, a fixed rational lattice, and density of the clock section. The explicit rates 1,sqrt(2),sqrt(3),sqrt(6) and radices 2,3,5,7 give rational span dimension four. This is an ordinary proof reviewed by AI agents, without Lean or independent human verification of the infinite theorem.

No rational independence of the original reciprocal prime logarithms is assumed or proved; the repeated running-LCM endpoint remains open.

independent_clock_channel_relations: not_applicable_without_exact_Lean_statement
The selected staircase matrix with ones on and below the diagonal and integer $r \geq 2$ above it has Smith factors 1,r-1,...,r-1. For the selected {2,3,5} kernel, any entrywise-integral diagonal clearing has determinant magnitude at least $4^{N - 1}$ , whereas the reduced rational determinant has numerator one up to sign. A rational single-target border has primitive polynomial plus or minus (tX-s), where s/t is the reduced Schur complement. These ordinary arithmetic arguments were reviewed by AI agents; they are not the existing Lean nonvanishing theorem and have no independent human review.

The optimality concerns diagonal clearing and one rational target. It does not rule out multi-target, nonlinear, recurrence, or new approximation methods and does not prove running-LCM irrationality.

selected_staircase_integral_height_boundary: not_applicable_without_exact_Lean_statement

The result statements, their limits and the counts are copied from the public Lean repository’s published records identified by b0ebeb9add157584. These records are copied into the page when the website is built. Introductory explanations and labels are written for this site. The diagram at the top is the exception: a program draws it from numbers the papers print and tests the numerical relationships shown. Its caption and explanation are written for this site.