Let $f (1) = f (2) = 1$ and for $n > 2$ $f (n) = f (n - f (n - 1)) + f (n - f (n - 2)) .$ Does $f (n)$ miss infinitely many integers? What is its behaviour?

Worked, still open.

number theory · open · formalized (Lean) · 0 attempts

use this record

vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

evidence

unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Your recurrence is the **Hofstadter $Q$-sequence** (OEIS **A005185**), sometimes called Hofstadter’s original “$Q$” or “$U$” sequence: [ Q(1)=Q(2)=1,\qquad Q(n)=Q\bigl(n-Q(n-1)\bigr)+Q\bigl(n-Q(n-2)\bigr)\ \ (n>2). ] ([OEIS][1])

candidate solution ↗

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.

candidate solution ↗

formal

AMS 11 · open (literature)

theorem erdos_422 : answer(sorry) ↔ Set.Infinite {n | ∀ x, f x ≠ n}

formal-conjectures/422.lean ↗

oeis

A005185 — Hofstadter Q-sequence: a(1) = a(2) = 1; a(n) = a(n-a(n-1)) + a(n-a(n-2)) for n > 2.1,1,2,3,3,4,5,5,6,6,6,8,8,8,10,9,10,11,11,12,12,12,12,16,14,14,16,16,16,16,20,17,17,20,21,19,20,22,21,22,23,23,24,24,24,

status

open

evidence

formal

oeis

status

Search Vela