The Three-Prime Running Least Common Multiple

Introduction

Let $P$ be a finite set of primes with $|P|\ge2$, and let $a_1<a_2<\cdots$ enumerate the positive integers all of whose prime factors lie in $P$. Erdős Problem #269 asks whether $$\sum_{n\ge1}\frac{1}{[a_1,\ldots,a_n]}$$ is irrational, where $[a_1,\ldots,a_n]$ is the least common multiple [1][2]. Numbering and current reported status follow Bloom’s catalogue, whose current record reproduces the displayed problem and explicitly warns that its OPEN label is the website owner’s present assessment and may omit relevant literature . The original publications, not the catalogue, carry the mathematical claims. The universal problem remains open, but this note settles every two-prime instance; unresolved finite cases begin with $|P|=3$. In the primary 1988 source, Erdős states the infinite-$P$ assertion as a simple exercise and presents persistence for a finite number of primes greater than one as a probable extension, not as a theorem; the current open status is supplied by the catalogue rather than inferred from that conjectural wording.

Three things are known and fix the shape of the question. The restriction $|P|\ge2$ is necessary: for $P=\{p\}$ the enumeration is $a_n=p^{\,n-1}$, so $[a_1,\ldots,a_n]=p^{\,n-1}$ and the sum is $p/(p-1)$, which is rational. For that reason we use the modern restriction $|P|\ge2$; the 1974 letter itself writes “given primes $p_1,\ldots,p_r$” without explicitly restricting $r$. For infinite $P$ the sum is always irrational, which Erdős calls a simple exercise . And in a letter of 1 January 1973 he recorded that he could prove irrationality once duplicate running-LCM values are removed [3]; the one-page letter states this result but does not include its proof.

That last remark is the one this note is closest to. The running least common multiple is not injective in $n$: it is constant along stretches of the enumeration and changes only at certain points. Removing duplicate summands means summing over the distinct values rather than over $n$. Section  proves that both this de-duplicated sum and the original repeated sum are transcendental when $|P|=2$, by reducing them to a Hecke–Mahler value whose transcendence goes back to Loxton and van der Poorten [10], quoted here in the modern form of Bugeaud and Laurent’s Theorem 1.1 [9]. This is an independent route, not a recovery of the unprinted argument in the letter, and it is not the first public proof: Steve Fan posted the same factorisation, the same Hecke–Mahler reduction, and the same conclusion in the discussion thread of the problem’s page on 26 June 2026 [11], with follow-up remarks there extending the argument to arbitrary coprime pairs. This manuscript was first released publicly on 22 July 2026, 26 days later, at commit a9d3ab8. Sections and  make the finite ingredients of the three-prime reindexing exact: they identify where the value is constant, by exactly what factor it changes when it changes, and give a finite rectangular-box fibre identity. They do not establish the ordered three-prime infinite passage and imply no irrationality result for three primes.

Throughout, $p,q,r$ are pairwise distinct primes, and a is a power $b^{e}$ of a single base; a is a power of $3$, and similarly for the other bases. Call $n$ when $n=p^{i}q^{j}r^{k}$ for some $i,j,k\ge0$; this is the . For $x\ge1$ write $$\operatorname{L}(x)=\operatorname{lcm}\{\,n\le x:\ n\ \text{smooth}\,\}, \qquad \operatorname{H}(x)=p^{\lfloor\log_p x\rfloor}\,q^{\lfloor\log_q x\rfloor}\, r^{\lfloor\log_r x\rfloor},$$ the and the , where $\lfloor\log_b x\rfloor$ is the integer logarithm, the largest $e$ with $b^{e}\le x$. Since $a_1,\ldots,a_n$ are exactly the smooth numbers up to $a_n$, we have $[a_1,\ldots,a_n]=\operatorname{L}(a_n)$, and the summands of the problem are the reciprocals of $\operatorname{L}$ along the enumeration. The reciprocal of the height at a smooth point is the $$\operatorname{K}(i,j,k)=\frac{1}{\operatorname{H}(p^{i}q^{j}r^{k})} .$$ For $b\in\{p,q,r\}$ we call the set of positive powers of $b$ the .

We treat $|P|=3$ throughout, writing $P=\{p,q,r\}$, and the smallest instance is $\{2,3,5\}$. Two of the statements proved here are unconditional and exact rather than approximate. Theorem  determines exactly the alphabet of the dyadic compression of the , that is, of the sequence of prime multipliers of $\operatorname{L}$ read in increasing order of the points where $\operatorname{L}$ increases (Section ): the radix $\beta_a$ of the block between $2^{a}$ and $2^{a+1}$, meaning the product of the multipliers that occur in that block, takes one of the four values $2$, $6$, $10$, $30$ and no others. Theorem  shows that the smallest two-by-two restriction of the kernel at $\{2,3,5\}$ has determinant $-1/15$ and hence rank two. In particular, it is not a product $f(i)g(j)h(k)$, so no argument may assume that form for the kernel on that rectangle. A rank-two matrix is itself a sum of two rank-one matrices, so decompositions into several separable terms are not excluded; what is excluded is the single product form. The mechanism is a small computation: $\operatorname{H}(6)=4\cdot3\cdot5=60$ and not $6$, because the running least common multiple at a smooth cutoff already sees powers of the other primes that the cutoff itself does not contain. The two-prime proof of Section  does not require such a separation.

The identification $\operatorname{L}(x)=\operatorname{H}(x)$ of Theorem  is what makes every statement about $\operatorname{L}$ below computable from three integer logarithms. In the unrestricted case that identification follows by iterating the prime-exponent maximum rule for least common multiples: $\operatorname{lcm}(1,\ldots,N)=\prod_{t\le N}t^{\lfloor\log_t N\rfloor}$, the product being over the primes $t\le N$ . Chebyshev’s function is the corresponding prime-power sum, so equivalently $\log\operatorname{lcm}(1,\ldots,N)=\psi(N)$  [6]; Montgomery–Vaughan state this exact identity directly in Exercise 6.2.7 [7]. We record the $P$-restricted form because every later statement in the formal development is derived from it, and we claim nothing new for it.

The line of argument developed below has four stages. First, identify $\operatorname{L}$ exactly as a product of three pure powers, so that every later statement is a statement about three integer logarithms. Second, compress its jump word into dyadic blocks, whose radix takes only the four values of Theorem . Third, assume the sum rational with reduced denominator $D$, factor $D=D_{\mathrm{sm}}B$ where every prime divisor of $D_{\mathrm{sm}}$ lies in $\{2,3,5\}$ and $\gcd(B,30)=1$, prove that the integral carry states, the integers the argument tracks from step to step (Section ), share the factor $D_{\mathrm{sm}}$, cancel it, and obtain a positive reduced carry $d_n$ bounded by a denominator-dependent $K(B,n)$ and satisfying $d_{n+1}=b_nd_n-Bm_n$. Fourth, find a stretch of consecutive steps over which the accumulated base and forcing put the residue of that carry above the bound, which is impossible. The first two stages and the finite core of the fourth are proved and formalised here. The algebraic absorption-and-cancellation core of the third stage is also checked, but the problem-specific claim that the actual carry shares $D_{\mathrm{sm}}$ is not. That rationality-to-carry instantiation and the cofinal existence of windows in the fourth stage are the two obligations of Section . The third and fourth stages follow a pattern standard in irrationality proofs; what is specific here is that the radix word driving the carry is the block alphabet of Theorem .

The statement of Problem #269 has been formalised before, as a conjecture with an unfilled proof, in the collection . That is a formal statement of the question up to a harmless rational normalisation: its Nat-indexed series includes the empty-prefix least-common-multiple term. Its rational, irrational, and infinite-prime assertions all end in sorry. The declarations described below are propositions about the objects the question is posed over. No claim of priority is made for any of these declarations. Kovač and Tao [8] treat several irrationality problems of Erdős for series of unit fractions by elementary means; nothing from that work is used here. We offer no numerical evidence about the value of the sum itself.

The table records two independent facts. The column says whether the statement is supported by an argument in this paper together with a linked Lean declaration, by a direct computation, or by no result in this paper. The last column states its logical reach and the nearest boundary. Thus a conditional implication may be checked in Lean while its hypothesis remains open, and the dyadic-window scan is a direct computation over $106{,}666$ instances rather than an unbounded theorem.

The running least common multiple as a product of pure powers

The smooth numbers up to $x$ are indexed by the exponent triples $(i,j,k)$ with $i\le\lfloor\log_p x\rfloor$, $j\le\lfloor\log_q x\rfloor$, $k\le\lfloor\log_r x\rfloor$ and $p^{i}q^{j}r^{k}\le x$: the coordinate bounds make the index set finite, and the last condition is the actual constraint. This is the . Both conditions matter: the coordinate box is strictly larger than the prefix, since a product of three large pure powers can exceed $x$ while each factor does not.

The identification below is routine, and its unrestricted analogue is classical, as recalled in the introduction; the short proof is given because every later statement is derived from it in the formal development.

Formalised as the , from the and the three membership statements for the pure powers, namely the , , and pure-power memberships.

The hypothesis that the primes are pairwise distinct is used exactly once, in the coprimality step, and it is not decorative: without it the three pure powers need not have coprime orders and their product need not divide the least common multiple. The identity is what makes every later statement about $\operatorname{L}$ computable from three integer logarithms.

At $(p,q,r)=(2,3,5)$ it reads $\operatorname{L}(x)=2^{\lfloor\log_2 x\rfloor}3^{\lfloor\log_3 x\rfloor} 5^{\lfloor\log_5 x\rfloor}$, whose first ten values are $$\begin{array}{c|cccccccccc} x&1&2&3&4&5&6&7&8&9&10\\ \hline \operatorname{L}(x)&1&2&6&12&60&60&60&120&360&360 \end{array}$$ So $\operatorname{L}(10)=8\cdot9\cdot5=360$, which is indeed the least common multiple of the smooth numbers $1,2,3,4,5,6,8,9,10$, and $\operatorname{L}(6)=4\cdot3\cdot5=60$ rather than $6$: the running value at a smooth cutoff already contains powers of the other two primes that the cutoff itself does not. These ten values also illustrate two of the statements of Section , since the value is constant on $\{5,6,7\}$ and on $\{9,10\}$, and each change multiplies by a single prime, by $2$ at $x=2,4,8$, by $3$ at $x=3,9$, and by $5$ at $x=5$.

Formalised as the , which needs no primality. The exponent $3$ is the number of generating primes, and the bound is the reason the reciprocal kernel is comparable to $x^{-3}$ rather than to $x^{-1}$.

Constancy on logarithmic cells and the jump points

By Theorem  the running value depends on $x$ only through the three integer logarithms. It is therefore constant wherever none of them changes and moves only where one of them does, and the next definition names the sets on which they are all constant, so that the next two theorems can say where the value stands still and by what factor it moves.

Say that $x$ and $y$ lie in the same when $\lfloor\log_b x\rfloor=\lfloor\log_b y\rfloor$ for each of $b=p,q,r$: the .

Formalised as the , the , and the . The running value therefore changes only when one of the three logarithms changes, and the next theorem says by exactly how much.

Formalised as the , , and coordinate steps, over the corresponding statements for the height alone, which need no primality: the , the , and the .

The jump points are therefore the pure prime powers, and they do not collide across channels.

Formalised as the and the , over the , the , and the ; the channels themselves are the . The exponent $0$ is omitted from each channel because it is the shared initial value, which is why the origin is counted once rather than three times.

Two channels never meet, so at each positive pure power exactly one of the three logarithms advances, and by Theorem  the running value is multiplied there by the prime of that channel. Reading those multipliers in increasing order of the pure powers gives the of $\operatorname{L}$: one letter from $\{p,q,r\}$ for each positive pure power, recording which prime the value is multiplied by at that point. At $\{2,3,5\}$ the pure powers in increasing order are $2,3,4,5,8,9,16,25,27,32,\ldots$, so the jump word begins $$2,\;3,2,\;5,2,\;3,2,\;5,3,2,\;\ldots$$ where the grouping is the one used in Section : each group ends at a power of two.

The two-prime sums are transcendental

The jump description gives a complete result in rank two. Temporarily let $P=\{p,q\}$ with $p<q$, and write $$L_{p,q}(t)=p^{\lfloor\log_p t\rfloor}q^{\lfloor\log_q t\rfloor}.$$ The argument of Theorem , with one coordinate omitted, identifies this as the running least common multiple of the $\{p,q\}$-smooth numbers up to $t$. Let $$\mathcal D_{p,q} =1+\sum_{t\in\{p,p^2,\ldots\}\cup\{q,q^2,\ldots\}} \frac1{L_{p,q}(t)} .$$ Thus $\mathcal D_{p,q}$ retains the initial value and exactly one reciprocal for every later distinct running-LCM value.

The same value also controls the series before repeated running-LCM values are removed. Because each $\{p,q\}$-smooth number is uniquely $p^iq^j$, that original series is $$\mathcal R_{p,q} =\sum_{i,j\ge0}\frac1{L_{p,q}(p^iq^j)}.$$

Thus both versions are settled, at the stronger level of transcendence, for $|P|=2$. A third prime replaces the single Beatty boundary by a genuinely two-dimensional ordering problem, so neither theorem supplies a three-prime irrationality result.

The height-fibre normal form

Grouping a lattice sum by the value of the running least common multiple turns it into a sum over heights whose coefficients are multiplicities. We prove this exactly, on a finite rectangular box.

Write $\mathcal B(h_p,h_q,h_r)$ for the box of exponent triples with $i\le h_p$, $j\le h_q$, $k\le h_r$, the , and, for a height $H$, write $F(H)$ for the set of points of the box whose running height $\operatorname{H}(p^{i}q^{j}r^{k})$ equals $H$, the . Distinct lattice points can carry the same height, and the coefficients of the normal form are exactly the sizes of these fibres.

Formalised as the , over the , with the height of a lattice point given by the .

For a worked instance take $(p,q,r)=(2,3,5)$ and the box $\mathcal B(1,1,1)$, whose eight points carry the smooth values $1,2,3,5,6,10,15,30$ and the heights $$\begin{array}{c|cccccccc} p^{i}q^{j}r^{k}&1&2&3&5&6&10&15&30\\ \hline \operatorname{H}&1&2&6&60&60&360&360&10800 \end{array}$$ Six heights occur, two of them twice: the points $5$ and $6$ share the height $60$, and $10$ and $15$ share the height $360$. Theorem  here reads $$1+\tfrac12+\tfrac16+\tfrac1{60}+\tfrac1{60}+\tfrac1{360}+\tfrac1{360} +\tfrac1{10800} =1+\tfrac12+\tfrac16+\tfrac2{60}+\tfrac2{360}+\tfrac1{10800} =\tfrac{18421}{10800}.$$ The two coefficients $2$ carry the whole content of the regrouping on this box; where the heights are pairwise distinct the identity is a relabelling and nothing more.

Together with Theorem  this is the finite core of the ordered prime-power jump expansion: the value is constant on cells, the cells are indexed by heights, and the coefficient of a height is the number of lattice points it collects. The passage to the infinite sum, and the explicit ordering of the pure powers that would make the expansion a series in the jumps, are not proved here. Theorem  is an identity between two finite sums, and it is stated over the full box rather than the smooth prefix, so it is not a statement about $\operatorname{L}$ at a cutoff.

The same module records a one-step map $\tau(b,d,s)=b(s-d)$, the , with the rewriting . No orbit of this map is analysed.

A quadratic bound for smooth exponent shells

A tail estimate needs to know how many lattice points can share a short multiplicative interval. Fix a box $\mathcal B(h_p,h_q,h_r)$ and an interval $[\lambda,\eta)$, and write $\mathcal S$ for the set of exponent triples of that box whose smooth value lies in that interval, the . Every bound below assumes the interval short in the multiplicative sense, meaning that $\eta$ is at most a stated multiple of $\lambda$.

Informally, the next lemma says that an interval whose right endpoint is at most $b$ times its left endpoint contains at most one of the numbers $b^{a}w$.

Formalised as the . The hypothesis is that the interval has multiplicative width at most $b$.

Formalised as the and the .

Formalised as the , over the . The constant $9$ is the square of the number of generating primes and appears because the bound is the arithmetic–geometric comparison for a sum of three sorted coordinates; sorting is a hypothesis, not a normalisation, since the shell itself is not symmetric in the three bases. The last inequality of the proof is an equality when $h_p=h_q=h_r$, both sides then being $9(h_p+1)^{2}$, so no constant larger than $9$ survives that step.

The bound is uniform in $\lambda$ and $\eta$ subject to the width condition, and it is stated for the actual filtered shell rather than for a lattice model of it. It is an input to a tail estimate and is not itself one: no series is bounded here. The estimate is elementary and uses no analytic input on the distribution of smooth numbers, only the projection of Proposition .

Non-separability of the three-prime kernel

One might hope to write the three-prime kernel as $f(i)g(j)h(k)$ and so reduce the problem to one-dimensional criteria. Such a factorisation would force the value at $(1,1,0)$ to be determined by the values at $(0,0,0)$, $(1,0,0)$ and $(0,1,0)$, since the four values would then satisfy $\operatorname{K}(0,0,0)\,\operatorname{K}(1,1,0)=\operatorname{K}(1,0,0)\,\operatorname{K}(0,1,0)$. Theorem  computes those four values at $\{2,3,5\}$ and finds that they do not: the failure is exact, and it occurs on the smallest rectangle on which it could occur.

Formalised as the , over the four exact values, the , , , and . The exact determinant calculation is the . The height at $6$ is $60$ rather than $6$ because the maximal pure powers below $6$ are $4$, $3$ and $5$: the running least common multiple at a smooth cutoff sees powers of the other primes that the cutoff itself does not contain. That is the mechanism behind the non-separation.

A nonzero two-by-two minor rules out writing the kernel as $f(i)g(j)h(k)$ on this box, so no argument may assume that single product form here. It rules out nothing further: a rank-two matrix is itself a sum of two rank-one matrices, so sums of separable terms, higher-rank decompositions, separable majorants, changes of variable, and one-dimensional estimates applied after a decomposition all remain available. It gives no rank lower bound beyond two, and it is not an independence or irrationality statement.

Dyadic blocks and a conditional carry contradiction

The section makes three moves and leaves one hypothesis standing. We first compress the jump word of Section  into the blocks cut out by consecutive powers of two, and show that the multiplier of a block takes only four values (Theorem ). We then record what can be cancelled from a hypothetical denominator, and exactly where that cancellation is still conditional (Theorem ). Finally we leave the smooth numbers behind and argue with integer sequences alone: for a multiplier coprime to $30$, no positive sequence obeying the cleared recurrence can stay inside its bound once a certain residue condition holds arbitrarily far out (Theorem ). That residue condition, condition  below, is the hypothesis; it is not proved here, and the closing subsection reports a finite computation, which is evidence for it and not a proof of it.

Complements and further questions

Theorems  and  close both two-prime questions, at the stronger level of transcendence. The three-prime de-duplicated and repeated series are outside the one-dimensional Hecke–Mahler reduction used in those theorems and remain open. At three primes the exact unresolved statement is best separated from the bridge and from the possible methods for proving it.

Guide to the formal sources

Each linked phrase opens its Lean declaration at the pinned source revision 76b5b0a7ed5d. The running-LCM structure, residue arithmetic, and local-window bridge occupy three separate modules. Four distinctions are worth carrying into the source. The height statements hold for arbitrary bases, while the statements about $\operatorname{L}$ need the three primes to be distinct. The normal form of Section  is a finite identity over a rectangular box, not a convergence theorem. The smooth part of a hypothetical denominator can be cancelled only after divisibility of the actual carry states by that factor is proved. And the formal cofinal-escape predicate is an unproved hypothesis of Theorem ; its application to the actual $\{2,3,5\}$ word is not asserted.