The board / number theory / #408
Erdős Problem #408
OPENdatabase updated 2025-08-31
No proof or disproof is known, and it isn't known whether a finite computation could settle it.
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`f n` is \(f(n) = \min\{k \geq 1 : \phi_k(n) = 1\}\), where \(\phi_1 = \phi\) is Euler's totient function and \(\phi_k(n) = \phi(\phi_{k-1}(n))\). As in the source, \(k\) starts at \(1\), so \(f(1) = 1\). Every iterate of \(\phi\) at \(0\) is \(0\), so `sInf ∅ = 0` gives the junk value \(f(0) = 0\).
Let \(\phi(n)\) be the Euler totient function and \(\phi_k(n)\) be the iterated \(\phi\) function, so that \(\phi_1(n)=\phi(n)\) and \(\phi_k(n)=\phi(\phi_{k-1}(n))\). Let \[f(n) = \min\{k : \phi_k(n)=1\}.\] Does \(f(n)/\log n\) have a distribution function? We ask for a distribution function \(F\) (non-decreasing, with \(F(-\infty) = 0\) and \(F(+\infty) = 1\)) such that, for every continuity point \(c\) of \(F\), the set \(\{n : f(n)/\log n \leq c\}\) has natural density \(F(c)\). For \(n \leq 1\) the quotient is \(0\) in Lean, which does not affect densities. Erdős, Granville, Pomerance and Spiro [EGPS90] proved that the answer is yes, conditional on a form of the Elliott–Halberstam conjecture.
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research open · 3 further variants / known partial results in the Lean file. The authoritative wording is on erdosproblems.com/408.Prizenone
Topicsnumber theory, iterated functions
Statement in LeanFormal Conjectures
Solution in Leanno
OEISA049108
Original pageerdosproblems.com/408
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Live page status OPEN with 0 claimed proofs, but “Currently working on this problem” lists Svyable; stopped to avoid colliding with the current worker.
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